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Chapter 8 Resource Masters

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Page 1: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Chapter 8Resource Masters

Page 2: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Consumable WorkbooksMany of the worksheets contained in the Chapter Resource Masters bookletsare available as consumable workbooks.

Study Guide and Intervention Workbook 0-07-828029-XSkills Practice Workbook 0-07-828023-0Practice Workbook 0-07-828024-9

ANSWERS FOR WORKBOOKS The answers for Chapter 8 of these workbookscan be found in the back of this Chapter Resource Masters booklet.

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved.Printed in the United States of America. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teacher, and families without charge; and be used solely in conjunction with Glencoe’s Algebra 2. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher.

Send all inquiries to:The McGraw-Hill Companies8787 Orion PlaceColumbus, OH 43240-4027

ISBN: 0-07-828011-7 Algebra 2Chapter 8 Resource Masters

1 2 3 4 5 6 7 8 9 10 066 11 10 09 08 07 06 05 04 03 02

Glencoe/McGraw-Hill

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© Glencoe/McGraw-Hill iii Glencoe Algebra 2

Contents

Vocabulary Builder . . . . . . . . . . . . . . . . vii

Lesson 8-1Study Guide and Intervention . . . . . . . . 455–456Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 457Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 458Reading to Learn Mathematics . . . . . . . . . . 459Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 460

Lesson 8-2Study Guide and Intervention . . . . . . . . 461–462Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 463Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 464Reading to Learn Mathematics . . . . . . . . . . 465Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 466

Lesson 8-3Study Guide and Intervention . . . . . . . . 467–468Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 469Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 470Reading to Learn Mathematics . . . . . . . . . . 471Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 472

Lesson 8-4Study Guide and Intervention . . . . . . . . 473–474Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 475Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 476Reading to Learn Mathematics . . . . . . . . . . 477Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 478

Lesson 8-5Study Guide and Intervention . . . . . . . 479–480Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 481Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 482Reading to Learn Mathematics . . . . . . . . . . 483Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 484

Lesson 8-6Study Guide and Intervention . . . . . . . . 485–486Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 487Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 488Reading to Learn Mathematics . . . . . . . . . . 489Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 490

Lesson 8-7Study Guide and Intervention . . . . . . . . 491–492Skills Practice . . . . . . . . . . . . . . . . . . . . . . . 493Practice . . . . . . . . . . . . . . . . . . . . . . . . . . . 494Reading to Learn Mathematics . . . . . . . . . . 495Enrichment . . . . . . . . . . . . . . . . . . . . . . . . . 496

Chapter 8 AssessmentChapter 8 Test, Form 1 . . . . . . . . . . . . 497–498Chapter 8 Test, Form 2A . . . . . . . . . . . 499–500Chapter 8 Test, Form 2B . . . . . . . . . . . 501–502Chapter 8 Test, Form 2C . . . . . . . . . . . 503–504Chapter 8 Test, Form 2D . . . . . . . . . . . 505–506Chapter 8 Test, Form 3 . . . . . . . . . . . . 507–508Chapter 8 Open-Ended Assessment . . . . . . 509Chapter 8 Vocabulary Test/Review . . . . . . . 510Chapter 8 Quizzes 1 & 2 . . . . . . . . . . . . . . . 511Chapter 8 Quizzes 3 & 4 . . . . . . . . . . . . . . . 512Chapter 8 Mid-Chapter Test . . . . . . . . . . . . 513Chapter 8 Cumulative Review . . . . . . . . . . . 514Chapter 8 Standardized Test Practice . . 515–516

Standardized Test Practice Student Recording Sheet . . . . . . . . . . . . . . A1

ANSWERS . . . . . . . . . . . . . . . . . . . . . . A2–A32

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© Glencoe/McGraw-Hill iv Glencoe Algebra 2

Teacher’s Guide to Using theChapter 8 Resource Masters

The Fast File Chapter Resource system allows you to conveniently file the resourcesyou use most often. The Chapter 8 Resource Masters includes the core materials neededfor Chapter 8. These materials include worksheets, extensions, and assessment options.The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing in theAlgebra 2 TeacherWorks CD-ROM.

Vocabulary Builder Pages vii–viiiinclude a student study tool that presentsup to twenty of the key vocabulary termsfrom the chapter. Students are to recorddefinitions and/or examples for each term.You may suggest that students highlight orstar the terms with which they are notfamiliar.

WHEN TO USE Give these pages tostudents before beginning Lesson 8-1.Encourage them to add these pages to theirAlgebra 2 Study Notebook. Remind them to add definitions and examples as theycomplete each lesson.

Study Guide and InterventionEach lesson in Algebra 2 addresses twoobjectives. There is one Study Guide andIntervention master for each objective.

WHEN TO USE Use these masters asreteaching activities for students who needadditional reinforcement. These pages canalso be used in conjunction with the StudentEdition as an instructional tool for studentswho have been absent.

Skills Practice There is one master foreach lesson. These provide computationalpractice at a basic level.

WHEN TO USE These masters can be used with students who have weakermathematics backgrounds or needadditional reinforcement.

Practice There is one master for eachlesson. These problems more closely followthe structure of the Practice and Applysection of the Student Edition exercises.These exercises are of average difficulty.

WHEN TO USE These provide additionalpractice options or may be used ashomework for second day teaching of thelesson.

Reading to Learn MathematicsOne master is included for each lesson. Thefirst section of each master asks questionsabout the opening paragraph of the lessonin the Student Edition. Additionalquestions ask students to interpret thecontext of and relationships among termsin the lesson. Finally, students are asked tosummarize what they have learned usingvarious representation techniques.

WHEN TO USE This master can be usedas a study tool when presenting the lessonor as an informal reading assessment afterpresenting the lesson. It is also a helpfultool for ELL (English Language Learner)students.

Enrichment There is one extensionmaster for each lesson. These activities mayextend the concepts in the lesson, offer anhistorical or multicultural look at theconcepts, or widen students’ perspectives onthe mathematics they are learning. Theseare not written exclusively for honorsstudents, but are accessible for use with alllevels of students.

WHEN TO USE These may be used asextra credit, short-term projects, or asactivities for days when class periods areshortened.

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© Glencoe/McGraw-Hill v Glencoe Algebra 2

Assessment OptionsThe assessment masters in the Chapter 8Resource Masters offer a wide range ofassessment tools for intermediate and finalassessment. The following lists describe eachassessment master and its intended use.

Chapter Assessment CHAPTER TESTS• Form 1 contains multiple-choice questions

and is intended for use with basic levelstudents.

• Forms 2A and 2B contain multiple-choicequestions aimed at the average levelstudent. These tests are similar in formatto offer comparable testing situations.

• Forms 2C and 2D are composed of free-response questions aimed at the averagelevel student. These tests are similar informat to offer comparable testingsituations. Grids with axes are providedfor questions assessing graphing skills.

• Form 3 is an advanced level test withfree-response questions. Grids withoutaxes are provided for questions assessinggraphing skills.

All of the above tests include a free-response Bonus question.

• The Open-Ended Assessment includesperformance assessment tasks that aresuitable for all students. A scoring rubricis included for evaluation guidelines.Sample answers are provided forassessment.

• A Vocabulary Test, suitable for allstudents, includes a list of the vocabularywords in the chapter and ten questionsassessing students’ knowledge of thoseterms. This can also be used in conjunc-tion with one of the chapter tests or as areview worksheet.

Intermediate Assessment• Four free-response quizzes are included

to offer assessment at appropriateintervals in the chapter.

• A Mid-Chapter Test provides an optionto assess the first half of the chapter. It iscomposed of both multiple-choice andfree-response questions.

Continuing Assessment• The Cumulative Review provides

students an opportunity to reinforce andretain skills as they proceed throughtheir study of Algebra 2. It can also beused as a test. This master includes free-response questions.

• The Standardized Test Practice offerscontinuing review of algebra concepts invarious formats, which may appear onthe standardized tests that they mayencounter. This practice includes multiple-choice, grid-in, and quantitative-comparison questions. Bubble-in andgrid-in answer sections are provided onthe master.

Answers• Page A1 is an answer sheet for the

Standardized Test Practice questionsthat appear in the Student Edition onpages 468–469. This improves students’familiarity with the answer formats theymay encounter in test taking.

• The answers for the lesson-by-lessonmasters are provided as reduced pageswith answers appearing in red.

• Full-size answer keys are provided forthe assessment masters in this booklet.

Page 6: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Reading to Learn MathematicsVocabulary Builder

NAME ______________________________________________ DATE ____________ PERIOD _____

88

© Glencoe/McGraw-Hill vii Glencoe Algebra 2

Voca

bula

ry B

uild

erThis is an alphabetical list of the key vocabulary terms you will learn in Chapter 8.As you study the chapter, complete each term’s definition or description. Rememberto add the page number where you found the term. Add these pages to your AlgebraStudy Notebook to review vocabulary at the end of the chapter.

Vocabulary Term Found on Page Definition/Description/Example

asymptote

A·suhm(p)·TOHT

center of a circle

center of an ellipse

circle

conic section

conjugate axis

KAHN·jih·guht

directrix

duh·REHK·trihks

distance formula

ellipse

ih·LIHPS

(continued on the next page)

Page 7: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill viii Glencoe Algebra 2

Vocabulary Term Found on Page Definition/Description/Example

foci of an ellipse

focus of a parabola

FOH·kuhs

hyperbola

hy·PUHR·buh·luh

latus rectum

LA·tuhs REHK·tuhm

major axis

midpoint formula

minor axis

parabola

puh·RA·buh·luh

tangent

TAN·juhnt

transverse axis

Reading to Learn MathematicsVocabulary Builder (continued)

NAME ______________________________________________ DATE ____________ PERIOD _____

88

Page 8: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Study Guide and InterventionMidpoint and Distance Formulas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

© Glencoe/McGraw-Hill 455 Glencoe Algebra 2

Less

on

8-1

The Midpoint Formula

Midpoint Formula The midpoint M of a segment with endpoints (x1, y1) and (x2, y2) is ! , ".y1 ! y2"2x1 ! x2"2

Find the midpoint of theline segment with endpoints at (4, !7) and (!2, 3).

! , " # ! , "# ! , " or (1, $2)

The midpoint of the segment is (1, $2).

$4"2

2"2

$7 ! 3"2

4 ! ($2)""2

y1 ! y2"2x1 ! x2"2

A diameter A!B! of a circlehas endpoints A(5, !11) and B(!7, 6).What are the coordinates of the centerof the circle?

The center of the circle is the midpoint of allof its diameters.

! , " # ! , "# ! , " or !$1, $2 "

The circle has center !$1, $2 ".1"2

1"2

$5"2

$2"2

$11 ! 6""2

5 ! ($7)""2

y1 ! y2"2

x1 ! x2"2

Example 1Example 1 Example 2Example 2

ExercisesExercises

Find the midpoint of each line segment with endpoints at the given coordinates.

1. (12, 7) and ($2, 11) 2. ($8, $3) and (10, 9) 3. (4, 15) and (10, 1)

(5, 9) (1, 3) (7, 8)

4. ($3, $3) and (3, 3) 5. (15, 6) and (12, 14) 6. (22, $8) and ($10, 6)

(0, 0) (13.5, 10) (6, !1)

7. (3, 5) and ($6, 11) 8. (8, $15) and ($7, 13) 9. (2.5, $6.1) and (7.9, 13.7)

"! , 8# " , !1# (5.2, 3.8)

10. ($7, $6) and ($1, 24) 11. (3, $10) and (30, $20) 12. ($9, 1.7) and ($11, 1.3)

(!4, 9) " , !15# (!10, 1.5)

13. Segment M#N# has midpoint P. If M has coordinates (14, $3) and P has coordinates ($8, 6), what are the coordinates of N? (!30, 15)

14. Circle R has a diameter S#T#. If R has coordinates ($4, $8) and S has coordinates (1, 4),what are the coordinates of T? (!9, !20)

15. Segment A#D# has midpoint B, and B#D# has midpoint C. If A has coordinates ($5, 4) and C has coordinates (10, 11), what are the coordinates of B and D?

B is "5, 8 #, D is "15, 13 #.1"

2"

33"

1"

3"

ODDS use formulas(show work)EVENS use programs
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© Glencoe/McGraw-Hill 456 Glencoe Algebra 2

The Distance Formula

Distance FormulaThe distance between two points (x1, y1) and (x2, y2) is given by d # $(x2 $#x1)2 !# (y2 $# y1)2#.

What is the distance between (8, !2) and (!6, !8)?

d # $(x2 $#x1)2 !#( y2 $# y1)2# Distance Formula

# $($6 $# 8)2 !# [$8 $# ($2)]#2# Let (x1, y1) # (8, $2) and (x2, y2) # ($6, $8).

# $($14)#2 ! ($#6)2# Subtract.

# $196 !# 36# or $232# Simplify.

The distance between the points is $232# or about 15.2 units.

Find the perimeter and area of square PQRS with vertices P(!4, 1),Q(!2, 7), R(4, 5), and S(2, !1).

Find the length of one side to find the perimeter and the area. Choose P#Q#.

d # $(x2 $#x1)2 !#( y2 $# y1)2# Distance Formula

# $[$4 $# ($2)]#2 ! (1# $ 7)2# Let (x1, y1) # ($4, 1) and (x2, y2) # ($2, 7).

# $($2)2#! ($6#)2# Subtract.

# $40# or 2$10# Simplify.

Since one side of the square is 2$10#, the perimeter is 8$10# units. The area is (2$10#)2, or40 units2.

Find the distance between each pair of points with the given coordinates.

1. (3, 7) and ($1, 4) 2. ($2, $10) and (10, $5) 3. (6, $6) and ($2, 0)

5 units 13 units 10 units4. (7, 2) and (4, $1) 5. ($5, $2) and (3, 4) 6. (11, 5) and (16, 9)

3$2! units 10 units $41! units7. ($3, 4) and (6, $11) 8. (13, 9) and (11, 15) 9. ($15, $7) and (2, 12)

3$34! units 2$10! units 5$26! units

10. Rectangle ABCD has vertices A(1, 4), B(3, 1), C($3, $2), and D($5, 1). Find theperimeter and area of ABCD. 2$13! # 6$5! units; 3$65! units2

11. Circle R has diameter S#T# with endpoints S(4, 5) and T($2, $3). What are thecircumference and area of the circle? (Express your answer in terms of %.)10$ units; 25$ units2

Study Guide and Intervention (continued)

Midpoint and Distance Formulas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

Example 1Example 1

Example 2Example 2

ExercisesExercises

ODDS use formulas(show work)EVENS use programs
Page 10: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Skills PracticeMidpoint and Distance Formulas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

© Glencoe/McGraw-Hill 457 Glencoe Algebra 2

Less

on

8-1

Find the midpoint of each line segment with endpoints at the given coordinates.

1. (4, $1), ($4, 1) (0, 0) 2. ($1, 4), (5, 2) (2, 3)

3. (3, 4), (5, 4) (4, 4) 4. (6, 2), (2, $1) "4, #

5. (3, 9), ($2, $3) " , 3# 6. ($3, 5), ($3, $8) "!3, ! #

7. (3, 2), ($5, 0) (!1, 1) 8. (3, $4), (5, 2) (4, !1)

9. ($5, $9), (5, 4) "0, ! # 10. ($11, 14), (0, 4) "! , 9#

11. (3, $6), ($8, $3) "! , ! # 12. (0, 10), ($2, $5) "!1, #

Find the distance between each pair of points with the given coordinates.

13. (4, 12), ($1, 0) 13 units 14. (7, 7), ($5, $2) 15 units

15. ($1, 4), (1, 4) 2 units 16. (11, 11), (8, 15) 5 units

17. (1, $6), (7, 2) 10 units 18. (3, $5), (3, 4) 9 units

19. (2, 3), (3, 5) $5! units 20. ($4, 3), ($1, 7) 5 units

21. ($5, $5), (3, 10) 17 units 22. (3, 9), ($2, $3) 13 units

23. (6, $2), ($1, 3) $74! units 24. ($4, 1), (2, $4) $61! units

25. (0, $3), (4, 1) 4$2! units 26. ($5, $6), (2, 0) $85! units

5"

9"

5"

11"

5"

3"

1"

1"

Page 11: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill 458 Glencoe Algebra 2

Find the midpoint of each line segment with endpoints at the given coordinates.

1. (8, $3), ($6, $11) (1, !7) 2. ($14, 5), (10, 6) "!2, #3. ($7, $6), (1, $2) (!3, !4) 4. (8, $2), (8, $8) (8, !5)

5. (9, $4), (1, $1) "5, ! # 6. (3, 3), (4, 9) " , 6#7. (4, $2), (3, $7) " , ! # 8. (6, 7), (4, 4) "5, #9. ($4, $2), ($8, 2) (!6, 0) 10. (5, $2), (3, 7) "4, #

11. ($6, 3), ($5, $7) "! , !2# 12. ($9, $8), (8, 3) "! , ! #13. (2.6, $4.7), (8.4, 2.5) (5.5, !1.1) 14. !$ , 6", ! , 4" " , 5#15. ($2.5, $4.2), (8.1, 4.2) (2.8, 0) 16. ! , ", !$ , $ " "! , 0#

Find the distance between each pair of points with the given coordinates.

17. (5, 2), (2, $2) 5 units 18. ($2, $4), (4, 4) 10 units

19. ($3, 8), ($1, $5) $173! units 20. (0, 1), (9, $6) $130! units

21. ($5, 6), ($6, 6) 1 unit 22. ($3, 5), (12, $3) 17 units

23. ($2, $3), (9, 3) $157! units 24. ($9, $8), ($7, 8) 2$65! units

25. (9, 3), (9, $2) 5 units 26. ($1, $7), (0, 6) $170! units

27. (10, $3), ($2, $8) 13 units 28. ($0.5, $6), (1.5, 0) 2$10! units

29. ! , ", !1, " 1 unit 30. ($4$2#, $$5#), ($5$2#, 4$5#) $127! units

31. GEOMETRY Circle O has a diameter A#B#. If A is at ($6, $2) and B is at ($3, 4), find thecenter of the circle and the length of its diameter. "! , 1#; 3$5! units

32. GEOMETRY Find the perimeter of a triangle with vertices at (1, $3), ($4, 9), and ($2, 1).18 # 2$17! units

9"

7"5

3"5

2"5

1"

1"2

5"8

1"2

1"8

1"

2"3

1"3

5"

1"

11"

5"

11"

9"

7"

7"

5"

11"

Practice (Average)

Midpoint and Distance Formulas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

ODDS onlyUse PROGRAMS!!!!
Page 12: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

Reading to Learn MathematicsMidpoint and Distance Formulas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

© Glencoe/McGraw-Hill 459 Glencoe Algebra 2

Less

on

8-1

Pre-Activity How are the Midpoint and Distance Formulas used in emergencymedicine?

Read the introduction to Lesson 8-1 at the top of page 412 in your textbook.

How do you find distances on a road map?

Sample answer: Use the scale of miles on the map. You mightalso use a ruler.

Reading the Lesson

1. a. Write the coordinates of the midpoint of a segment with endpoints (x1, y1) and (x2, y2).

" , #b. Explain how to find the midpoint of a segment if you know the coordinates of the

endpoints. Do not use subscripts in your explanation.

Sample answer: To find the x-coordinate of the midpoint, add the x-coordinates of the endpoints and divide by two. To find the y-coordinate of the midpoint, do the same with the y-coordinates ofthe endpoints.

2. a. Write an expression for the distance between two points with coordinates (x1, y1) and(x2, y2). $(x2 !!x1)2 #! (y2 !! y1)2!

b. Explain how to find the distance between two points. Do not use subscripts in yourexplanation.

Sample answer: Find the difference between the x-coordinates and square it. Find the difference between the y-coordinates and square it. Add the squares. Then find the squareroot of the sum.

3. Consider the segment connecting the points ($3, 5) and (9, 11).

a. Find the midpoint of this segment. (3, 8)b. Find the length of the segment. Write your answer in simplified radical form. 6$5!

Helping You Remember

4. How can the “mid” in midpoint help you remember the midpoint formula?

Sample answer: The midpoint is the point in the middle of a segment. Itis halfway between the endpoints. The coordinates of the midpoint arefound by finding the average of the two x-coordinates (add them anddivide by 2) and the average of the two y-coordinates.

y1 # y2"2x1 # x2"2

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© Glencoe/McGraw-Hill 460 Glencoe Algebra 2

Quadratic FormConsider two methods for solving the following equation.

(y $ 2)2 $ 5(y $ 2) ! 6 # 0

One way to solve the equation is to simplify first, then use factoring.

y2 $ 4y ! 4 $ 5y ! 10 ! 6 # 0y2 $ 9y ! 20 # 0

( y $ 4)( y $ 5) # 0

Thus, the solution set is {4, 5}.

Another way to solve the equation is first to replace y $ 2 by a single variable.This will produce an equation that is easier to solve than the original equation.Let t # y $ 2 and then solve the new equation.

( y $ 2)2 $ 5( y $ 2) ! 6 # 0t2 $ 5t ! 6 # 0

(t $ 2)(t $ 3) # 0

Thus, t is 2 or 3. Since t # y $ 2, the solution set of the original equation is {4, 5}.

Solve each equation using two different methods.

1. (z ! 2)2 ! 8(z ! 2) ! 7 # 0 2. (3x $ 1)2 $ (3x $ 1) $ 20 # 0

3. (2t ! 1)2 $ 4(2t ! 1) ! 3 # 0 4. ( y2 $ 1)2 $ ( y2 $ 1) $ 2 # 0

5. (a2 $ 2)2 $ 2(a2 $ 2) $ 3 # 0 6. (1 ! $c#)2 ! (1 ! $c#) $ 6 # 0

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

8-18-1

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Study Guide and InterventionParabolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-28-2

© Glencoe/McGraw-Hill 461 Glencoe Algebra 2

Less

on

8-2

Equations of Parabolas A parabola is a curve consisting of all points in thecoordinate plane that are the same distance from a given point (the focus) and a given line(the directrix). The following chart summarizes important information about parabolas.

Standard Form of Equation y # a(x $ h)2 ! k x # a(y $ k)2 ! h

Axis of Symmetry x # h y # k

Vertex (h, k ) (h, k )

Focus !h, k ! " !h ! , k"Directrix y # k $ x # h $

Direction of Opening upward if a & 0, downward if a ' 0 right if a & 0, left if a ' 0

Length of Latus Rectum units units

Identify the coordinates of the vertex and focus, the equations ofthe axis of symmetry and directrix, and the direction of opening of the parabolawith equation y % 2x2 ! 12x ! 25.

y # 2x2 $ 12x $ 25 Original equationy # 2(x2 $ 6x) $ 25 Factor 2 from the x-terms.y # 2(x2 $ 6x ! ■) $ 25 $ 2(■) Complete the square on the right side.y # 2(x2 $ 6x ! 9) $ 25 $ 2(9) The 9 added to complete the square is multiplied by 2.y # 2(x $ 3)2 $ 43 Write in standard form.

The vertex of this parabola is located at (3, $43), the focus is located at !3, $42 ", the

equation of the axis of symmetry is x # 3, and the equation of the directrix is y # $43 .The parabola opens upward.

Identify the coordinates of the vertex and focus, the equations of the axis ofsymmetry and directrix, and the direction of opening of the parabola with thegiven equation.

1. y # x2 ! 6x $ 4 2. y # 8x $ 2x2 ! 10 3. x # y2 $ 8y ! 6

(!3, !13), (2, 18), "2, 17 #, (!10, 4), "!9 , 4#,"!3, !12 #, x % !3, x % 2, y % 18 , y % 4, x % !10 ,

y % !13 , up down right

Write an equation of each parabola described below.

4. focus ($2, 3), directrix x # $2 5. vertex (5, 1), focus !4 , 1"x % 6(y ! 3)2 ! 2 x % !3(y ! 1)2 # 51

"

11"12

1"12

1"

1"

1"

3"

3"

1"

1"8

7"8

1"a

1"a

1"4a

1"4a

1"4a

1"4a

ExampleExample

ExercisesExercises

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© Glencoe/McGraw-Hill 462 Glencoe Algebra 2

Graph Parabolas To graph an equation for a parabola, first put the given equation instandard form.

y # a(x $ h)2 ! k for a parabola opening up or down, orx # a(y $ k)2 ! h for a parabola opening to the left or right

Use the values of a, h, and k to determine the vertex, focus, axis of symmetry, and length ofthe latus rectum. The vertex and the endpoints of the latus rectum give three points on theparabola. If you need more points to plot an accurate graph, substitute values for pointsnear the vertex.

Graph y % (x ! 1)2 # 2.

In the equation, a # , h # 1, k # 2.

The parabola opens up, since a & 0.vertex: (1, 2)axis of symmetry: x # 1

focus: !1, 2 ! " or !1, 2 "

length of latus rectum: or 3 units

endpoints of latus rectum: !2 , 2 ", !$ , 2 "

The coordinates of the focus and the equation of the directrix of a parabola aregiven. Write an equation for each parabola and draw its graph.

1. (3, 5), y # 1 2. (4, $4), y# $6 3. (5, $1), x # 3

y % (x ! 3)2 # 3 y % (x ! 4)2 ! 5 x % (y # 1)2 # 41"

1"

1"

x

y

Ox

y

O

x

y

O

3"4

1"2

3"4

1"2

1"

"13"

3"4

1"4!"

13""

x

y

O

1"3

1"3

Study Guide and Intervention (continued)

Parabolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-28-2

ExampleExample

ExercisesExercises

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Skills PracticeParabolas

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8-28-2

© Glencoe/McGraw-Hill 463 Glencoe Algebra 2

Less

on

8-2

Write each equation in standard form.

1. y # x2 ! 2x ! 2 2. y # x2 $ 2x ! 4 3. y # x2 ! 4x ! 1

y % [x ! (!1)]2 # 1 y % (x ! 1)2 # 3 y % [x ! (!2)]2 # (!3)

Identify the coordinates of the vertex and focus, the equations of the axis ofsymmetry and directrix, and the direction of opening of the parabola with thegiven equation. Then find the length of the latus rectum and graph the parabola.

4. y # (x $ 2)2 5. x # (y $ 2)2 ! 3 6. y # $(x ! 3)2 ! 4

vertex: (2, 0); vertex: (3, 2); vertex: (!3, 4); focus: "2, #; focus: "3 , 2#; focus: "!3, 3 #;axis of symmetry: axis of symmetry: axis of symmetry: x % 2; y % 2; x % !3;directrix: y % ! ; directrix: x % 2 ; directrix: y % 4 ; opens up; opens right; opens down;latus rectum: 1 unit latus rectum: 1 unit latus rectum: 1 unit

Write an equation for each parabola described below. Then draw the graph.

7. vertex (0, 0), 8. vertex (5, 1), 9. vertex (1, 3),

focus !0, $ " focus !5, " directrix x #

y % !3x2 y % (x ! 5)2 # 1 x % 2(y ! 3)2 # 1

x

y

Ox

y

O

x

y

O

7"8

5"4

1"12

1"

3"

1"

3"

1"

1"

x

y

Ox

y

O

x

y

O

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Write each equation in standard form.

1. y # 2x2 $ 12x ! 19 2. y # x2 ! 3x ! 3. y # $3x2 $ 12x $ 7

y % 2(x ! 3)2 # 1 y % [x ! (!3)]2 # (!4) y % !3[x ! (!2)]2 # 5Identify the coordinates of the vertex and focus, the equations of the axis ofsymmetry and directrix, and the direction of opening of the parabola with thegiven equation. Then find the length of the latus rectum and graph the parabola.

4. y # (x $ 4)2 ! 3 5. x # $ y2 ! 1 6. x # 3(y ! 1)2 $ 3

vertex: (4, 3); vertex: (1, 0); vertex: (!3, !1); focus: "4, 3 #; focus: " , 0#; focus: "!2 , !1#;axis: x % 4; axis: y % 0; axis: y % !1;directrix: y % 2 ; directrix: x % 1 ; directrix: x % !3 ; opens up; opens left; opens right;latus rectum: 1 unit latus rectum: 3 units latus rectum: unit

Write an equation for each parabola described below. Then draw the graph.

7. vertex (0, $4), 8. vertex ($2, 1), 9. vertex (1, 3),

focus !0, $3 " directrix x # $3 axis of symmetry x # 1,latus rectum: 2 units,a ' 0

y % 2x2 ! 4 x % (y ! 1)2 ! 2 y % ! (x ! 1)2 # 3

10. TELEVISION Write the equation in the form y # ax2 for a satellite dish. Assume that thebottom of the upward-facing dish passes through (0, 0) and that the distance from thebottom to the focus point is 8 inches. y % x21

"

x

y

Ox

y

O

1"

1"

7"8

1"

1"

3"

3"

11"

1"

1"

x

y

O

x

y

O

1"3

1"

1"2

1"2

Practice (Average)

Parabolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-28-2

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Reading to Learn MathematicsParabolas

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© Glencoe/McGraw-Hill 465 Glencoe Algebra 2

Less

on

8-2

Pre-Activity How are parabolas used in manufacturing?

Read the introduction to Lesson 8-2 at the top of page 419 in your textbook.

Name at least two reflective objects that might have the shape of aparabola.

Sample answer: telescope mirror, satellite dish

Reading the Lesson

1. In the parabola shown in the graph, the point (2, $2) is called

the and the point (2, 0) is called the

. The line y # $4 is called the

, and the line x # 2 is called the

.

2. a. Write the standard form of the equation of a parabola that opens upward ordownward. y % a(x ! h)2 # k

b. The parabola opens downward if and opens upward if . The

equation of the axis of symmetry is , and the coordinates of the vertex are

.

3. A parabola has equation x # $ ( y $ 2)2 ! 4. This parabola opens to the .

It has vertex and focus . The directrix is . The length

of the latus rectum is units.

Helping You Remember

4. How can the way in which you plot points in a rectangular coordinate system help you toremember what the sign of a tells you about the direction in which a parabola opens?Sample answer: In plotting points, a positive x-coordinate tells you tomove to the right and a negative x-coordinate tells you to move to theleft. This is like a parabola whose equation is of the form “x % …”; itopens to the right if a & 0 and to the left if a ' 0. Likewise, a positive y-coordinate tells you to move up and a negative y-coordinate tells youto move down. This is like a parabola whose equation is of the form “y % …”; it opens upward if a & 0 and downward if a ' 0.

8x % 6(2, 2)(4, 2)

left1"8

(h, k)x % h

a & 0a ' 0

axis of symmetrydirectrix

focusvertex

x

y

O

(2, –2)

(2, 0)

y % –4

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Tangents to ParabolasA line that intersects a parabola in exactly one point without crossing the curve is a tangent to the parabola. The point where a tangent line touches a parabola is the point of tangency. The line perpendicular to a tangent to a parabola at the point of tangency is called the normal to the parabola at that point. In the diagram, line ! is tangent to the

parabola that is the graph of y # x2 at !"32", "

94"". The

x-axis is tangent to the parabola at O, and the y-axis is the normal to the parabola at O.

Solve each problem.

1. Find an equation for line ! in the diagram. Hint: A nonvertical line with anequation of the form y # mx ! b will be tangent to the graph of y # x2 at

!"32", "

94"" if and only if !"

32", "

94"" is the only pair of numbers that satisfies both

y # x2 and y # mx ! b.

2. If a is any real number, then (a, a2) belongs to the graph of y # x2. Express m and b in terms of a to find an equation of the form y # mx ! b for the linethat is tangent to the graph of y # x2 at (a, a2).

3. Find an equation for the normal to the graph of y # x2 at !"32", "

94"".

4. If a is a nonzero real number, find an equation for the normal to the graph ofy # x2 at (a, a2).

x

y

O

!

y % x2

1–1–2–3 2

6

5

4

3

2

1

3

!3–2, 9–4"

Enrichment

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8-28-2

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Study Guide and InterventionCircles

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8-38-3

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8-3

Equations of Circles The equation of a circle with center (h, k) and radius r units is (x $ h)2 ! (y $ k)2 # r2.

Write an equation for a circle if the endpoints of a diameter are at(!4, 5) and (6, !3).

Use the midpoint formula to find the center of the circle.

(h, k) # ! , " Midpoint formula

# ! , " (x1, y1) # ($4, 5), (x2, y2) # (6, $3)

# ! , " or (1, 1) Simplify.

Use the coordinates of the center and one endpoint of the diameter to find the radius.

r # $(x2 $x#1)2 !#( y2 $# y1)2# Distance formula

r # $($4 $# 1)2 !# (5 $#1)2# (x1, y1) # (1, 1), (x2, y2) # ($4, 5)

# $($5)2# ! 42# # $41# Simplify.

The radius of the circle is $41#, so r2 # 41.

An equation of the circle is (x $ 1)2 ! (y $ 1)2 # 41.

Write an equation for the circle that satisfies each set of conditions.

1. center (8, $3), radius 6 (x ! 8)2 # (y # 3)2 % 36

2. center (5, $6), radius 4 (x ! 5)2 # (y # 6)2 % 16

3. center ($5, 2), passes through ($9, 6) (x # 5)2 # (y ! 2)2 % 32

4. endpoints of a diameter at (6, 6) and (10, 12) (x ! 8)2 # (y ! 9)2 % 13

5. center (3, 6), tangent to the x-axis (x ! 3)2 # (y ! 6)2 % 36

6. center ($4, $7), tangent to x # 2 (x # 4)2 # (y # 7)2 % 36

7. center at ($2, 8), tangent to y # $4 (x # 2)2 # (y ! 8)2 % 144

8. center (7, 7), passes through (12, 9) (x ! 7)2 # (y ! 7)2 % 29

9. endpoints of a diameter are ($4, $2) and (8, 4) (x ! 2)2 # (y ! 1)2 % 45

10. endpoints of a diameter are ($4, 3) and (6, $8) (x ! 1)2 # (y # 2.5)2 % 55.25

2"2

2"2

5 ! ($3)""2

$4 ! 6"2

y1 ! y2"2

x1 ! x2"2

ExampleExample

ExercisesExercises

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Graph Circles To graph a circle, write the given equation in the standard form of theequation of a circle, (x $ h)2 ! (y $ k)2 # r2.

Plot the center (h, k) of the circle. Then use r to calculate and plot the four points (h ! r, k),(h $ r, k), (h, k ! r), and (h, k $ r), which are all points on the circle. Sketch the circle thatgoes through those four points.

Find the center and radius of the circle whose equation is x2 # 2x # y2 # 4y % 11. Then graph the circle.

x2 ! 2x ! y2 ! 4y # 11x2 ! 2x ! ■ ! y2 ! 4y ! ■ # 11 !■

x2 ! 2x ! 1 ! y2 ! 4y ! 4 # 11 ! 1 ! 4(x ! 1)2 ! ( y ! 2)2 # 16

Therefore, the circle has its center at ($1, $2) and a radius of $16# # 4. Four points on the circle are (3, $2), ($5, $2), ($1, 2),and ($1, $6).

Find the center and radius of the circle with the given equation. Then graph thecircle.

1. (x $ 3)2 ! y2 # 9 2. x2 ! (y ! 5)2 # 4 3. (x $ 1)2 ! (y ! 3)2 # 9

(3, 0), r % 3 (0, !5), r % 2 (1, !3), r % 3

4. (x $ 2)2 ! (y ! 4)2 # 16 5. x2 ! y2 $ 10x ! 8y ! 16 # 0 6. x2 ! y2 $ 4x ! 6y # 12

(2, !4), r % 4 (5, !4), r % 5 (2, !3), r % 5

x

y

Ox

y

Ox

y

O

x

y

Ox

y

O

x

y

O

x

y

O

x2 # 2x # y2 # 4y % 11

Study Guide and Intervention (continued)

Circles

NAME ______________________________________________ DATE ____________ PERIOD _____

8-38-3

ExampleExample

ExercisesExercises

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Skills PracticeCircles

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8-38-3

© Glencoe/McGraw-Hill 469 Glencoe Algebra 2

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on

8-3

Write an equation for the circle that satisfies each set of conditions.

1. center (0, 5), radius 1 unit 2. center (5, 12), radius 8 unitsx2 # (y ! 5)2 % 1 (x ! 5)2 # (y ! 12)2 % 64

3. center (4, 0), radius 2 units 4. center (2, 2), radius 3 units(x ! 4)2 # y2 % 4 (x ! 2)2 # (y ! 2)2 % 9

5. center (4, $4), radius 4 units 6. center ($6, 4), radius 5 units(x ! 4)2 # (y # 4)2 % 16 (x # 6)2 # (y ! 4)2 % 25

7. endpoints of a diameter at ($12, 0) and (12, 0) x2 # y2 % 1448. endpoints of a diameter at ($4, 0) and ($4, $6) (x # 4)2 # (y # 3)2 % 99. center at (7, $3), passes through the origin (x ! 7)2 # (y # 3)2 % 58

10. center at ($4, 4), passes through ($4, 1) (x # 4)2 # (y ! 4)2 % 911. center at ($6, $5), tangent to y-axis (x # 6)2 # (y # 5)2 % 3612. center at (5, 1), tangent to x-axis (x ! 5)2 # (y ! 1)2 % 1

Find the center and radius of the circle with the given equation. Then graph thecircle.

13. x2 ! y2 # 9 14. (x $ 1)2 ! (y $ 2)2 # 4 15. (x ! 1)2 ! y2 # 16

(0, 0), 3 units (1, 2), 2 units (!1, 0), 4 units

16. x2 ! (y ! 3)2 # 81 17. (x $ 5)2 ! (y ! 8)2 # 49 18. x2 ! y2 $ 4y $ 32 # 0

(0, !3), 9 units (5, !8), 7 units (0, 2), 6 units

x

y

O 4 8

8

4

–4

–8

–4–8

x

y

O 4 8 12

–4

–8

–12

x

y

O 6 12

12

6

–6

–12

–6–12

x

y

Ox

y

Ox

y

O

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Write an equation for the circle that satisfies each set of conditions.

1. center ($4, 2), radius 8 units 2. center (0, 0), radius 4 units(x # 4)2 # (y ! 2)2 % 64 x2 # y2 % 16

3. center !$ , $$3#", radius 5$2# units 4. center (2.5, 4.2), radius 0.9 unit

"x # #2 # (y # $3!)2 % 50 (x ! 2.5)2 # (y ! 4.2)2 % 0.815. endpoints of a diameter at ($2, $9) and (0, $5) (x # 1)2 # (y # 7)2 % 56. center at ($9, $12), passes through ($4, $5) (x # 9)2 # (y # 12)2 % 747. center at ($6, 5), tangent to x-axis (x # 6)2 # (y ! 5)2 % 25

Find the center and radius of the circle with the given equation. Then graph thecircle.

8. (x ! 3)2 ! y2 # 16 9. 3x2 ! 3y2 # 12 10. x2 ! y2 ! 2x ! 6y # 26(!3, 0), 4 units (0, 0), 2 units (!1, !3), 6 units

11. (x $ 1)2 ! y2 ! 4y # 12 12. x2 $ 6x ! y2 # 0 13. x2 ! y2 ! 2x ! 6y # $1(1, !2), 4 units (3, 0), 3 units (!1, !3), 3 units

WEATHER For Exercises 14 and 15, use the following information.On average, the circular eye of a hurricane is about 15 miles in diameter. Gale winds canaffect an area up to 300 miles from the storm’s center. In 1992, Hurricane Andrew devastatedsouthern Florida. A satellite photo of Andrew’s landfall showed the center of its eye on onecoordinate system could be approximated by the point (80, 26).

14. Write an equation to represent a possible boundary of Andrew’s eye.(x ! 80)2 # (y ! 26)2 % 56.25

15. Write an equation to represent a possible boundary of the area affected by gale winds.(x ! 80)2 # (y ! 26)2 % 90,000

x

y

O 4 8

4

–4

–8

–4–8x

y

Ox

y

O

1"

1"4

Practice (Average)

Circles

NAME ______________________________________________ DATE ____________ PERIOD _____

8-38-3

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Reading to Learn MathematicsCircles

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8-38-3

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Less

on

8-3

Pre-Activity Why are circles important in air traffic control?

Read the introduction to Lesson 8-3 at the top of page 426 in your textbook.

A large home improvement chain is planning to enter a new metropolitanarea and needs to select locations for its stores. Market research has shownthat potential customers are willing to travel up to 12 miles to shop at oneof their stores. How can circles help the managers decide where to placetheir store?

Sample answer: A store will draw customers who live inside acircle with center at the store and a radius of 12 miles. The management should select locations for whichas many people as possible live within a circle of radius 12 miles around one of the stores.

Reading the Lesson

1. a. Write the equation of the circle with center (h, k) and radius r.(x ! h)2 # (y ! k)2 % r 2

b. Write the equation of the circle with center (4, $3) and radius 5.(x ! 4)2 # (y # 3)2 % 25

c. The circle with equation (x ! 8)2 ! y2 # 121 has center and radius

.

d. The circle with equation (x $ 10)2 ! ( y ! 10)2 # 1 has center and

radius .

2. a. In order to find center and radius of the circle with equation x2 ! y2 ! 4x $ 6y $3 # 0,

it is necessary to . Fill in the missing parts of thisprocess.

x2 ! y2 ! 4x $ 6y $ 3 # 0

x2 ! y2 ! 4x $ 6y #

x2 ! 4x ! ! y2 $ 6y ! # ! !

(x ! )2 ! ( y $ )2 #

b. This circle has radius 4 and center at .

Helping You Remember

3. How can the distance formula help you to remember the equation of a circle?Sample answer: Write the distance formula. Replace (x1, y1) with (h, k)and (x2, y2) with (x, y). Replace d with r. Square both sides. Now youhave the equation of a circle.

(!2, 3)1632

943943

complete the square

1(10, !10)

11(!8, 0)

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Tangents to CirclesA line that intersects a circle in exactly one point is a tangent to the circle. In the diagram, line ! is tangent to the circle with equation x2 ! y2 # 25 at the point whose coordinates are (3, 4).

A line is tangent to a circle at a point P on the circle if and only if the line is perpendicular to the radius from the center of the circle to point P. This fact enables you to find an equation of the tangent to a circle at a point P if you know an equation for the circle and the coordinates of P.

Use the diagram above to solve each problem.

1. What is the slope of the radius to the point with coordinates (3, 4)? What isthe slope of the tangent to that point?

2. Find an equation of the line ! that is tangent to the circle at (3, 4).

3. If k is a real number between $5 and 5, how many points on the circle have x-coordinate k? State the coordinates of these points in terms of k.

4. Describe how you can find equations for the tangents to the points you namedfor Exercise 3.

5. Find an equation for the tangent at ($3, 4).

5

–5

–5

5

(3, 4)

y

xO

!x2 # y2 % 25

Enrichment

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8-38-3

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Study Guide and InterventionEllipses

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Less

on

8-4

Equations of Ellipses An ellipse is the set of all points in a plane such that the sumof the distances from two given points in the plane, called the foci, is constant. An ellipsehas two axes of symmetry which contain the major and minor axes. In the table, thelengths a, b, and c are related by the formula c2 # a2 $ b2.

Standard Form of Equation ! # 1 ! # 1

Center (h, k) (h, k)

Direction of Major Axis Horizontal Vertical

Foci (h ! c, k ), (h $ c, k ) (h, k $ c), (h, k ! c)

Length of Major Axis 2a units 2a units

Length of Minor Axis 2b units 2b units

Write an equation for the ellipse shown.

The length of the major axis is the distance between ($2, $2) and ($2, 8). This distance is 10 units.

2a # 10, so a # 5The foci are located at ($2, 6) and ($2, 0), so c # 3.

b2 # a2 $ c2

# 25 $ 9# 16

The center of the ellipse is at ($2, 3), so h # $2, k # 3,a2 # 25, and b2 # 16. The major axis is vertical.

An equation of the ellipse is ! # 1.

Write an equation for the ellipse that satisfies each set of conditions.

1. endpoints of major axis at ($7, 2) and (5, 2), endpoints of minor axis at ($1, 0) and ($1, 4)

# % 1

2. major axis 8 units long and parallel to the x-axis, minor axis 2 units long, center at ($2, $5)

# (y # 5)2 % 1

3. endpoints of major axis at ($8, 4) and (4, 4), foci at ($3, 4) and ($1, 4)

# % 1

4. endpoints of major axis at (3, 2) and (3, $14), endpoints of minor axis at ($1, $6) and (7, $6)

# % 1

5. minor axis 6 units long and parallel to the x-axis, major axis 12 units long, center at (6, 1)

# % 1(x ! 6)2"9

(y ! 1)2"36

(x ! 3)2"16

(y # 6)2"64

(y ! 4)2"35

(x # 2)2"36

(x # 2)2"16

(y ! 2)2"4

(x # 1)2"36

(x ! 2)2"16

( y $ 3)2"25

x

F1

F2O

y

(x $ h)2"

b2(y $ k)2"

a2(y $ k)2"

b2(x $ h)2"

a2

ExampleExample

ExercisesExercises

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© Glencoe/McGraw-Hill 474 Glencoe Algebra 2

Graph Ellipses To graph an ellipse, if necessary, write the given equation in thestandard form of an equation for an ellipse.

! # 1 (for ellipse with major axis horizontal) or

! # 1 (for ellipse with major axis vertical)

Use the center (h, k) and the endpoints of the axes to plot four points of the ellipse. To makea more accurate graph, use a calculator to find some approximate values for x and y thatsatisfy the equation.

Graph the ellipse 4x2 # 6y2 # 8x ! 36y % !34.

4x2 ! 6y2 ! 8x $ 36y # $344x2 ! 8x ! 6y2 $ 36y # $ 34

4(x2 ! 2x ! ■) ! 6( y2 $ 6y ! ■) # $34 ! ■4(x2 ! 2x ! 1) ! 6( y2 $ 6y ! 9) # $34 ! 58

4(x ! 1)2 ! 6( y $ 3)2 # 24

! # 1

The center of the ellipse is ($1, 3). Since a2 # 6, a # $6#.Since b2 # 4, b # 2.The length of the major axis is 2$6#, and the length of the minor axis is 4. Since the x-termhas the greater denominator, the major axis is horizontal. Plot the endpoints of the axes.Then graph the ellipse.

Find the coordinates of the center and the lengths of the major and minor axesfor the ellipse with the given equation. Then graph the ellipse.

1. ! # 1 (0, 0), 4$3!, 6 2. ! # 1 (0, 0), 10, 4

3. x2 ! 4y2 ! 24y # $32 (0, !3), 4, 2 4. 9x2 ! 6y2 $ 36x ! 12y # 12 (2, !1), 6, 2$6!

x

y

Ox

y

O

x

y

Ox

y

O

y2"4

x2"25

x2"9

y2"12

( y $ 3)2"4

(x ! 1)2"6 xO

y

4x2 # 6y2 # 8x ! 36y % !34

(x $ h)2"

b2( y $ k)2"

a2

( y $ k)2"

b2(x $ h)2"

a2

Study Guide and Intervention (continued)

Ellipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

ExampleExample

ExercisesExercises

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Skills PracticeEllipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

© Glencoe/McGraw-Hill 475 Glencoe Algebra 2

Less

on

8-4

Write an equation for each ellipse.

1. 2. 3.

# % 1 # % 1 # % 1

Write an equation for the ellipse that satisfies each set of conditions.

4. endpoints of major axis 5. endpoints of major axis 6. endpoints of major axis at (0, 6) and (0, $6), at (2, 6) and (8, 6), at (7, 3) and (7, 9),endpoints of minor axis endpoints of minor axis endpoints of minor axis at ($3, 0) and (3, 0) at (5, 4) and (5, 8) at (5, 6) and (9, 6)

# % 1 # % 1 # % 1

7. major axis 12 units long 8. endpoints of major axis 9. endpoints of major axis atand parallel to x-axis, at ($6, 0) and (6, 0), foci (0, 12) and (0, $12), foci atminor axis 4 units long, at ($$32#, 0) and ($32#, 0) (0, $23# ) and (0, $$23# )center at (0, 0)

# % 1 # % 1 # % 1

Find the coordinates of the center and foci and the lengths of the major andminor axes for the ellipse with the given equation. Then graph the ellipse.

10. ! # 1 11. ! # 1 12. ! # 1

(0, 0); (0, ($19!); (0, 0); ((6$2!, 0); (0, 0), (0, (2$6!); 20; 18 18; 6 14; 10

x

y

O 4 8

8

4

–4

–8

–4–8x

y

O 4 8

8

4

–4

–8

–4–8x

y

O 4 8

8

4

–4

–8

–4–8

x2"25

y2"49

y2"9

x2"81

x2"81

y2"100

x2"

y2"

y2"

x2"

y2"

x2"

(x ! 7)2"4

(y ! 6)2"9

(y ! 6)2"4

(x ! 5)2"9

x2"

y2"

(y ! 2)2"9

x2"

x2"

y2"

y2"

x2"

xO

y(0, 5)

(0, –1)

(–4, 2) (4, 2)

xO

y

(0, 3)

(0, –3)

(0, –5)

(0, 5)

xO

y

(0, 2)

(0, –2)

(–3, 0)(3, 0)

Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
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© Glencoe/McGraw-Hill 476 Glencoe Algebra 2

Write an equation for each ellipse.

1. 2. 3.

# % 1 # % 1 # % 1

Write an equation for the ellipse that satisfies each set of conditions.

4. endpoints of major axis 5. endpoints of major axis 6. major axis 20 units long at ($9, 0) and (9, 0), at (4, 2) and (4, $8), and parallel to x-axis,endpoints of minor axis endpoints of minor axis minor axis 10 units long,at (0, 3) and (0, $3) at (1, $3) and (7, $3) center at (2, 1)

# % 1 # % 1 # % 1

7. major axis 10 units long, 8. major axis 16 units long, 9. endpoints of minor axis minor axis 6 units long center at (0, 0), foci at at (0, 2) and (0, $2), foci and parallel to x-axis, (0, 2$15# ) and (0, $2$15# ) at ($4, 0) and (4, 0)center at (2, $4)

# % 1 # % 1 # % 1

Find the coordinates of the center and foci and the lengths of the major andminor axes for the ellipse with the given equation. Then graph the ellipse.

10. ! # 1 11. ! # 1 12. ! # 1

(0, 0); (0, ($7!); 8; 6 (3, 1); (3, 1 ( $35! ); (!4, !3); 12; 2 (!4 ( 2$6!, !3); 14;

10

13. SPORTS An ice skater traces two congruent ellipses to form a figure eight. Assume that thecenter of the first loop is at the origin, with the second loop to its right. Write an equationto model the first loop if its major axis (along the x-axis) is 12 feet long and its minoraxis is 6 feet long. Write another equation to model the second loop.

x

y

O 4 8

8

4

–4

–8

–4–8x

y

O

( y ! 3)2"25

(x ! 4)2"49

(x $ 3)2"1

( y $ 1)2"36

x2"9

y2"16

y2"

x2"

x2"

y2"

(x ! 2)2"9

(y # 4)2"25

(y ! 1)2"25

(x ! 2)2"100

(x ! 4)2"9

(y # 3)2"25

y2"

x2"

(y ! 3)2"9

(x # 1)2"25

x2"

(y ! 2)2"9

y2"

x2"

xO

y

(–5, 3)

(–6, 3)

(3, 3)

(4, 3)

xO

y

(0, 2 ! $%5)

(0, 2 # $%5)

(0, –1)

(0, 5)

xO

y(0, 3)

(0, –3)

(–11, 0) (11, 0)6 12

2

–2

–6–12

Practice (Average)

Ellipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

Brad and Anissa Stuckey
Brad and Anissa Stuckey
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Reading to Learn MathematicsEllipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

© Glencoe/McGraw-Hill 477 Glencoe Algebra 2

Less

on

8-4

Pre-Activity Why are ellipses important in the study of the solar system?

Read the introduction to Lesson 8-4 at the top of page 433 in your textbook.

Is the Earth always the same distance from the Sun? Explain your answerusing the words circle and ellipse. No; if the Earth’s orbit were acircle, it would always be the same distance from the Sunbecause every point on a circle is the same distance from thecenter. However, the Earth’s orbit is an ellipse, and the pointson an ellipse are not all the same distance from the center.

Reading the Lesson1. An ellipse is the set of all points in a plane such that the of the

distances from two fixed points is . The two fixed points are called the

of the ellipse.

2. Consider the ellipse with equation ! # 1.

a. For this equation, a # and b # .

b. Write an equation that relates the values of a, b, and c. c2 % a2 ! b2

c. Find the value of c for this ellipse. $5!

3. Consider the ellipses with equations ! # 1 and ! # 1. Complete the

following table to describe characteristics of their graphs.

Standard Form of Equation ! # 1 ! # 1

Direction of Major Axis vertical horizontal

Direction of Minor Axis horizontal vertical

Foci (0, 3), (0, !3) ($5!, 0), (!$5!, 0)Length of Major Axis 10 units 6 units

Length of Minor Axis 8 units 4 units

Helping You Remember4. Some students have trouble remembering the two standard forms for the equation of an

ellipse. How can you remember which term comes first and where to place a and b inthese equations? The x-axis is horizontal. If the major axis is horizontal, the first term is . The y-axis is vertical. If the major axis is vertical, the

first term is . a is always the larger of the numbers a and b.y2"

x2"

y2"4

x2"9

x2"16

y2"25

y2"4

x2"9

x2"16

y2"25

23

y2"4

x2"9

fociconstant

sum

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© Glencoe/McGraw-Hill 478 Glencoe Algebra 2

Eccentricity In an ellipse, the ratio "d

c" is called the eccentricity and is denoted by the

letter e. Eccentricity measures the elongation of an ellipse. The closer e is to 0,the more an ellipse looks like a circle. The closer e is to 1, the more elongated

it is. Recall that the equation of an ellipse is "ax2

2" ! "by2

2" # 1 or "bx2

2" ! "ay2

2" # 1

where a is the length of the major axis, and that c # $a2 $ b#2#.

Find the eccentricity of each ellipse rounded to the nearesthundredth.

1. "x92" ! "3

y6

2" # 1 2. "8

x12" ! "

y9

2" # 1 3. "

x42" ! "

y9

2" # 1

0.87 0.94 0.75

4. "1x62" ! "

y9

2" # 1 5. "3

x62" ! "1

y6

2" # 1 6. "

x42" ! "3

y6

2" # 1

0.66 0.75 0.94

7. Is a circle an ellipse? Explain your reasoning.

Yes; it is an ellipse with eccentricity 0.

8. The center of the sun is one focus of Earth's orbit around the sun. Thelength of the major axis is 186,000,000 miles, and the foci are 3,200,000miles apart. Find the eccentricity of Earth's orbit.

approximately 0.17

9. An artificial satellite orbiting the earth travels at an altitude that variesbetween 132 miles and 583 miles above the surface of the earth. If thecenter of the earth is one focus of its elliptical orbit and the radius of theearth is 3950 miles, what is the eccentricity of the orbit?

approximately 0.052

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

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Study Guide and InterventionHyperbolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

© Glencoe/McGraw-Hill 479 Glencoe Algebra 2

Less

on

8-5

Equations of Hyperbolas A hyperbola is the set of all points in a plane such thatthe absolute value of the difference of the distances from any point on the hyperbola to anytwo given points in the plane, called the foci, is constant.

In the table, the lengths a, b, and c are related by the formula c2 # a2 ! b2.

Standard Form of Equation $ # 1 $ # 1

Equations of the Asymptotes y $ k # ( (x $ h) y $ k # ( (x $ h)

Transverse Axis Horizontal Vertical

Foci (h $ c, k), (h ! c, k) (h, k $ c), (h, k ! c)

Vertices (h $ a, k), (h ! a, k) (h, k $ a), (h, k ! a)

Write an equation for the hyperbola with vertices (!2, 1) and (6, 1)and foci (!4, 1) and (8, 1).

Use a sketch to orient the hyperbola correctly. The center of the hyperbola is the midpoint of the segment joining the two

vertices. The center is ( , 1), or (2, 1). The value of a is the

distance from the center to a vertex, so a # 4. The value of c is the distance from the center to a focus, so c # 6.

c2 # a2 ! b2

62 # 42 ! b2

b2 # 36 $ 16 # 20

Use h, k, a2, and b2 to write an equation of the hyperbola.

$ # 1

Write an equation for the hyperbola that satisfies each set of conditions.

1. vertices ($7, 0) and (7, 0), conjugate axis of length 10 ! % 1

2. vertices ($2, $3) and (4, $3), foci ($5, $3) and (7, $3) ! % 1

3. vertices (4, 3) and (4, $5), conjugate axis of length 4 ! % 1

4. vertices ($8, 0) and (8, 0), equation of asymptotes y # ( x ! % 1

5. vertices ($4, 6) and ($4, $2), foci ($4, 10) and ($4, $6) ! % 1(x # 4)2"48

(y ! 2)2"16

9y2"

x2"

1"6

(x ! 4)2"4

(y # 1)2"16

(y # 3)2"27

(x ! 1)2"9

y2"

x2"

( y $ 1)2"20

(x $ 2)2"16

$2 ! 6"2

x

y

O

a"b

b"a

(x $ h)2"

b2(y $ k)2"

a2(y $ k)2"

b2(x $ h)2"

a2

ExampleExample

ExercisesExercises

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© Glencoe/McGraw-Hill 480 Glencoe Algebra 2

Graph Hyperbolas To graph a hyperbola, write the given equation in the standardform of an equation for a hyperbola

$ # 1 if the branches of the hyperbola open left and right, or

$ # 1 if the branches of the hyperbola open up and down

Graph the point (h, k), which is the center of the hyperbola. Draw a rectangle withdimensions 2a and 2b and center (h, k). If the hyperbola opens left and right, the verticesare (h $ a, k) and (h ! a, k). If the hyperbola opens up and down, the vertices are (h, k $ a)and (h, k ! a).

Draw the graph of 6y2 ! 4x2 ! 36y ! 8x % !26.

Complete the squares to get the equation in standard form.6y2 $ 4x2 $ 36y $ 8x # $266( y2 $ 6y ! ■) $ 4(x2 ! 2x ! ■) # $26 ! ■6( y2 $ 6y ! 9) $ 4(x2 ! 2x ! 1) # $26 ! 506( y $ 3)2 $ 4(x ! 1)2 # 24

$ # 1

The center of the hyperbola is ($1, 3).According to the equation, a2 # 4 and b2 # 6, so a # 2 and b # $6#.The transverse axis is vertical, so the vertices are ($1, 5) and ($1, 1). Draw a rectangle withvertical dimension 4 and horizontal dimension 2$6# % 4.9. The diagonals of this rectangleare the asymptotes. The branches of the hyperbola open up and down. Use the vertices andthe asymptotes to sketch the hyperbola.

Find the coordinates of the vertices and foci and the equations of the asymptotesfor the hyperbola with the given equation. Then graph the hyperbola.

1. $ # 1 2. ( y $ 3)2 $ # 1 3. $ # 1

(2, 0), (!2, 0); (!2, 4), (!2, 2); (0, 4), (0, !4); (2$5!, 0), (!2$5!, 0); (!2, 3 # $10! ), (0, 5), (0, !5); y % (2x (!2, 3 ! $10! ); y % ( x

y % x # 3 ,

y % ! x # 2 1"

1"

2"

1"

xO

y

4"

x2"9

y2"16

(x ! 2)2"9

y2"16

x2"4

(x ! 1)2"6

( y $ 3)2"4 xO

y

(x $ h)2"

b2( y $ k)2"

a2

( y $ k)2""

b2(x $ h)2"

a2

Study Guide and Intervention (continued)

Hyperbolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

ExampleExample

ExercisesExercises

xO

y

xO

y

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Skills PracticeHyperbolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

© Glencoe/McGraw-Hill 481 Glencoe Algebra 2

Less

on

8-5

Write an equation for each hyperbola.

1. 2. 3.

! % 1 ! % 1 ! % 1

Write an equation for the hyperbola that satisfies each set of conditions.

4. vertices ($4, 0) and (4, 0), conjugate axis of length 8 ! % 1

5. vertices (0, 6) and (0, $6), conjugate axis of length 14 ! % 1

6. vertices (0, 3) and (0, $3), conjugate axis of length 10 ! % 1

7. vertices ($2, 0) and (2, 0), conjugate axis of length 4 ! % 1

8. vertices ($3, 0) and (3, 0), foci ((5, 0) ! % 1

9. vertices (0, 2) and (0, $2), foci (0, (3) ! % 1

10. vertices (0, $2) and (6, $2), foci (3 ( $13#, $2) ! % 1

Find the coordinates of the vertices and foci and the equations of the asymptotesfor the hyperbola with the given equation. Then graph the hyperbola.

11. $ # 1 12. $ # 1 13. $ # 1

((3, 0); ((3$5!, 0); (0, (7); (0, ($58! ); ((4, 0); (($17!, 0);y % (2x y % ( x y % ( x

xO

y

4 8

8

4

–4

–8

–4–8xO

y

4 8

8

4

–4

–8

–4–8xO

y

1"

7"

y2"1

x2"16

x2"9

y2"49

y2"36

x2"9

(y # 2)2"4

(x ! 3)2"9

x2"

y2"

y2"

x2"

y2"

x2"

x2"

y2"

x2"

y2"

y2"

x2"

y2"

x2"

x2"

y2"

y2"

x2"

x

y

O

($!29, 0)(–$!29, 0)

(2, 0)(–2, 0)

4 8

8

4

–4

–8

–4–8x

y

O

(0, $!61)

(0, –$!61)

(0, 6)

(0, –6)

4 8

8

4

–4

–8

–4–8x

y

O

($!41, 0)(–$!41, 0)

(5, 0)

(–5, 0)

4 8

8

4

–4

–8

–4–8

Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
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© Glencoe/McGraw-Hill 482 Glencoe Algebra 2

Write an equation for each hyperbola.

1. 2. 3.

! % 1 ! % 1 ! % 1

Write an equation for the hyperbola that satisfies each set of conditions.

4. vertices (0, 7) and (0, $7), conjugate axis of length 18 units ! % 1

5. vertices (1, $1) and (1, $9), conjugate axis of length 6 units ! % 1

6. vertices ($5, 0) and (5, 0), foci (($26#, 0) ! % 1

7. vertices (1, 1) and (1, $3), foci (1, $1 ( $5#) ! % 1

Find the coordinates of the vertices and foci and the equations of the asymptotesfor the hyperbola with the given equation. Then graph the hyperbola.

8. $ # 1 9. $ # 1 10. $ # 1

(0, (4); (0, (2$5!); (1, 3), (1, 1); (3, 0), (3, !4); y % (2x (1, 2 ( $5!); (3, !2 ( 2$2!);

y ! 2 % ( (x ! 1) y # 2 % ((x ! 3)

11. ASTRONOMY Astronomers use special X-ray telescopes to observe the sources ofcelestial X rays. Some X-ray telescopes are fitted with a metal mirror in the shape of ahyperbola, which reflects the X rays to a focus. Suppose the vertices of such a mirror arelocated at ($3, 0) and (3, 0), and one focus is located at (5, 0). Write an equation thatmodels the hyperbola formed by the mirror. ! % 1y2

"x2"

xO

y

xO

y

4 8

8

4

–4

–8

–4–8

1"

(x $ 3)2"4

( y ! 2)2"4

(x $ 1)2"4

( y $ 2)2"1

x2"4

y2"16

(x ! 1)2"1

(y # 1)2"4

y2"

x2"

(x ! 1)2"9

(y # 5)2"16

x2"

y2"

(y # 2)2"16

(x ! 1)2"4

(x # 3)2"25

(y ! 2)2"9

x2"

y2"

x

y

O(–1, –2)

(1, –2)

(3, –2)x

y

O

(–3, 2 # $!34)

(–3, 2 ! $!34)

(–3, –1)(–3, 5)

4

8

4

–4

–4–8x

y

O

(0, 3$%5)

(0, –3$%5)

(0, 3)

(0, –3)

4 8

8

4

–4

–8

–4–8

Practice (Average)

Hyperbolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
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Reading to Learn MathematicsHyperbolas

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

© Glencoe/McGraw-Hill 483 Glencoe Algebra 2

Less

on

8-5

Pre-Activity How are hyperbolas different from parabolas?

Read the introduction to Lesson 8-5 at the top of page 441 in your textbook.

Look at the sketch of a hyperbola in the introduction to this lesson. Listthree ways in which hyperbolas are different from parabolas.Sample answer: A hyperbola has two branches, while aparabola is one continuous curve. A hyperbola has two foci,while a parabola has one focus. A hyperbola has two vertices,while a parabola has one vertex.

Reading the Lesson

1. The graph at the right shows the hyperbola whose

equation in standard form is $ # 1.

The point (0, 0) is the of the hyperbola.

The points (4, 0) and ($4, 0) are the of the hyperbola.

The points (5, 0) and ($5, 0) are the of the hyperbola.

The segment connecting (4, 0) and ($4, 0) is called the axis.

The segment connecting (0, 3) and (0, $3) is called the axis.

The lines y # x and y # $ x are called the .

2. Study the hyperbola graphed at the right.

The center is .

The value of a is .

The value of c is .

To find b2, solve the equation # ! .

The equation in standard form for this hyperbola is .

Helping You Remember

3. What is an easy way to remember the equation relating the values of a, b, and c for ahyperbola? This equation looks just like the Pythagorean Theorem,although the variables represent different lengths in a hyperbola than ina right triangle.

"x42" ! "1

y22" % 1

b2a2c242

(0, 0)

x

y

O

asymptotes3"4

3"4

conjugatetransverse

foci

vertices

center

y2"9

x2"16

x

y

O(–4, 0) (4, 0)(–5, 0) (5, 0)

y % 34xy % – 34x

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© Glencoe/McGraw-Hill 484 Glencoe Algebra 2

Rectangular Hyperbolas A rectangular hyperbola is a hyperbola with perpendicular asymptotes.For example, the graph of x2 $ y2 # 1 is a rectangular hyperbola. A hyperbolawith asymptotes that are not perpendicular is called a nonrectangularhyperbola. The graphs of equations of the form xy # c, where c is a constant,are rectangular hyperbolas.

Make a table of values and plot points to graph each rectangularhyperbola below. Be sure to consider negative values for thevariables. See students’ tables.1. xy # $4 2. xy # 3

3. xy # $1 4. xy # 8

5. Make a conjecture about the asymptotes of rectangular hyperbolas.

The coordinate axes are the asymptotes.

x

y

Ox

y

O

x

y

Ox

y

O

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

8-58-5

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Study Guide and InterventionConic Sections

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

© Glencoe/McGraw-Hill 485 Glencoe Algebra 2

Less

on

8-6

Standard Form Any conic section in the coordinate plane can be described by anequation of the form

Ax2 ! Bxy ! Cy2 ! Dx ! Ey ! F # 0, where A, B, and C are not all zero.One way to tell what kind of conic section an equation represents is to rearrange terms andcomplete the square, if necessary, to get one of the standard forms from an earlier lesson.This method is especially useful if you are going to graph the equation.

Write the equation 3x2 ! 4y2 ! 30x ! 8y # 59 % 0 in standard form.State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola.

3x2 $ 4y2 $ 30x $ 8y ! 59 # 0 Original equation3x2 $ 30x $ 4y2 $ 8y # $59 Isolate terms.

3(x2 $ 10x ! ■) $ 4( y2 ! 2y ! ■) # $59 ! ■ ! ■ Factor out common multiples.3(x2 $ 10x ! 25) $ 4( y2 ! 2y ! 1) # $59 ! 3(25) ! ($4)(1) Complete the squares.

3(x $ 5)2 $ 4( y ! 1)2 # 12 Simplify.

$ # 1 Divide each side by 12.

The graph of the equation is a hyperbola with its center at (5, $1). The length of the transverse axis is 4 units and the length of the conjugate axis is 2$3# units.

Write each equation in standard form. State whether the graph of the equation isa parabola, circle, ellipse, or hyperbola.

1. x2 ! y2 $ 6x ! 4y ! 3 # 0 2. x2 ! 2y2 ! 6x $ 20y ! 53 # 0

(x ! 3)2 # (y # 2)2 % 10; circle # % 1; ellipse

3. 6x2 $ 60x $ y ! 161 # 0 4. x2 ! y2 $ 4x $14y ! 29 # 0

y % 6(x ! 5)2 # 11; parabola (x ! 2)2 # (y ! 7)2 % 24; circle

5. 6x2 $ 5y2 ! 24x ! 20y $ 56 # 0 6. 3y2 ! x $ 24y ! 46 # 0

! % 1; hyperbola x % !3(y ! 4)2 # 2; parabola

7. x2 $ 4y2 $ 16x ! 24y $ 36 # 0 8. x2 ! 2y2 ! 8x ! 4y ! 2 # 0

! % 1; hyperbola # % 1; ellipse

9. 4x2 ! 48x ! y ! 158 # 0 10. 3x2 ! y2 $ 48x $ 4y ! 184 # 0

y % !4(x # 6)2 ! 14; parabola # % 1; ellipse

11. $3x2 ! 2y2 $ 18x ! 20y ! 5 # 0 12. x2 ! y2 ! 8x ! 2y ! 8 # 0

! % 1; hyperbola (x # 4)2 # (y # 1)2 % 9; circle(x # 3)2"6

(y # 5)2"9

(y ! 2)2"12

(x ! 8)2"4

(y # 1)2"8

(x # 4)2"16

(y ! 3)2"16

(x ! 8)2"64

(y ! 2)2"12

(x # 2)2"10

(y ! 5)2"3

(x # 3)2"6

( y ! 1)2"3

(x $ 5)2"4

ExampleExample

ExercisesExercises

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© Glencoe/McGraw-Hill 486 Glencoe Algebra 2

Identify Conic Sections If you are given an equation of the formAx2 ! Bxy ! Cy2 ! Dx ! Ey ! F # 0, with B # 0,

you can determine the type of conic section just by considering the values of A and C. Referto the following chart.

Relationship of A and C Type of Conic Section

A # 0 or C # 0, but not both. parabola

A # C circle

A and C have the same sign, but A ) C. ellipse

A and C have opposite signs. hyperbola

Without writing the equation in standard form, state whether thegraph of each equation is a parabola, circle, ellipse, or hyperbola.

Study Guide and Intervention (continued)

Conic Sections

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

ExampleExample

a. 3x2 ! 3y2 # 5x # 12 % 0A # 3 and C # $3 have opposite signs, sothe graph of the equation is a hyperbola.

b. y2 % 7y ! 2x # 13A # 0, so the graph of the equation isa parabola.

ExercisesExercises

Without writing the equation in standard form, state whether the graph of eachequation is a parabola, circle, ellipse, or hyperbola.

1. x2 # 17x $ 5y ! 8 2. 2x2 ! 2y2 $ 3x ! 4y # 5parabola circle

3. 4x2 $ 8x # 4y2 $ 6y ! 10 4. 8(x $ x2) # 4(2y2 $ y) $ 100hyperbola circle

5. 6y2 $ 18 # 24 $ 4x2 6. y # 27x $ y2

ellipse parabola7. x2 # 4( y $ y2) ! 2x $ 1 8. 10x $ x2 $ 2y2 # 5y

ellipse ellipse9. x # y2 $ 5y ! x2 $ 5 10. 11x2 $ 7y2 # 77

circle hyperbola11. 3x2 ! 4y2 # 50 ! y2 12. y2 # 8x $ 11

circle parabola13. 9y2 $ 99y # 3(3x $ 3x2) 14. 6x2 $ 4 # 5y2 $ 3

circle hyperbola15. 111 # 11x2 ! 10y2 16. 120x2 $ 119y2 ! 118x $ 117y # 0

ellipse hyperbola17. 3x2 # 4y2 ! 12 18. 150 $ x2 # 120 $ y

hyperbola parabola

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Skills PracticeConic Sections

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

© Glencoe/McGraw-Hill 487 Glencoe Algebra 2

Less

on

8-6

Write each equation in standard form. State whether the graph of the equation isa parabola, circle, ellipse, or hyperbola. Then graph the equation.

1. x2 $ 25y2 # 25 hyperbola 2. 9x2 ! 4y2 # 36 ellipse 3. x2 ! y2 $ 16 # 0 circle! % 1 # % 1 x2 # y2 % 16

4. x2 ! 8x ! y2 # 9 circle 5. x2 ! 2x $ 15 # y parabola 6. 100x2 ! 25y2 # 400ellipse(x # 4)2 # y2 % 25 y % (x # 1)2 ! 16 # % 1

Without writing the equation in standard form, state whether the graph of eachequation is a parabola, circle, ellipse, or hyperbola.

7. 9x2 ! 4y2 # 36 ellipse 8. x2 ! y2 # 25 circle

9. y # x2 ! 2x parabola 10. y # 2x2 $ 4x $ 4 parabola

11. 4y2 $ 25x2 # 100 hyperbola 12. 16x2 ! y2 # 16 ellipse

13. 16x2 $ 4y2 # 64 hyperbola 14. 5x2 ! 5y2 # 25 circle

15. 25y2 ! 9x2 # 225 ellipse 16. 36y2 $ 4x2 # 144 hyperbola

17. y # 4x2 $ 36x $ 144 parabola 18. x2 ! y2 $ 144 # 0 circle

19. (x ! 3)2 ! ( y $ 1)2 # 4 circle 20. 25y2 $ 50y ! 4x2 # 75 ellipse

21. x2 $ 6y2 ! 9 # 0 hyperbola 22. x # y2 ! 5y $ 6 parabola

23. (x ! 5)2 ! y2 # 10 circle 24. 25x2 ! 10y2 $ 250 # 0 ellipse

x

y

O

xy

O 4 8

–4

–8

–12

–16

–4–8

x

y

O 4 8

8

4

–4

–8

–4–8

y2"

x2"

x

y

Ox

y

OxO

y

4 8

4

2

–2

–4

–4–8

y2"

x2"

y2"

x2"

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© Glencoe/McGraw-Hill 488 Glencoe Algebra 2

Write each equation in standard form. State whether the graph of the equation isa parabola, circle, ellipse, or hyperbola. Then graph the equation.

1. y2 # $3x 2. x2 ! y2 ! 6x # 7 3. 5x2 $ 6y2 $ 30x $ 12y # $9parabola circle hyperbolax % ! y 2 (x # 3)2 # y2 % 16 ! % 1

4. 196y2 # 1225 $ 100x2 5. 3x2 # 9 $ 3y2 $ 6y 6. 9x2 ! y2 ! 54x $ 6y # $81ellipse circle ellipse

# % 1 x2 # (y # 1)2 % 4 # % 1

Without writing the equation in standard form, state whether the graph of eachequation is a parabola, circle, ellipse, or hyperbola.

7. 6x2 ! 6y2 # 36 8. 4x2 $ y2 # 16 9. 9x2 ! 16y2 $ 64y $ 80 # 0 circle hyperbola ellipse

10. 5x2 ! 5y2 $ 45 # 0 11. x2 ! 2x # y 12. 4y2 $ 36x2 ! 4x $ 144 # 0circle parabola hyperbola

13. ASTRONOMY A satellite travels in an hyperbolic orbit. It reaches the vertex of its orbit

at (5, 0) and then travels along a path that gets closer and closer to the line y # x.

Write an equation that describes the path of the satellite if the center of its hyperbolicorbit is at (0, 0).

! % 1y2"

x2"

2"5

(y ! 3)2"9

(x # 3)2"1

y2"

x2"

xO

y

x

y

Ox

y

O

(y # 1)2"5

(x ! 3)2"6

1"

Practice (Average)

Conic Sections

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

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Reading to Learn MathematicsConic Sections

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

© Glencoe/McGraw-Hill 489 Glencoe Algebra 2

Less

on

8-6

Pre-Activity How can you use a flashlight to make conic sections?

Read the introduction to Lesson 8-6 at the top of page 449 in your textbook.

The figures in the introduction show how a plane can slice a double cone toform the conic sections. Name the conic section that is formed if the planeslices the double cone in each of the following ways:

• The plane is parallel to the base of the double cone and slices throughone of the cones that form the double cone. circle

• The plane is perpendicular to the base of the double cone and slicesthrough both of the cones that form the double cone. hyperbola

Reading the Lesson

1. Name the conic section that is the graph of each of the following equations. Give thecoordinates of the vertex if the conic section is a parabola and of the center if it is acircle, an ellipse, or a hyperbola.

a. ! # 1 ellipse; (3, !5)

b. x # $2( y ! 1)2 ! 7 parabola; (7, !1)c. (x $ 5)2 $ ( y ! 5)2 # 1 hyperbola; (5, !5)d. (x ! 6)2 ! ( y $ 2)2 # 1 circle; (!6, 2)

2. Each of the following is the equation of a conic section. For each equation, identify thevalues of A and C. Then, without writing the equation in standard form, state whetherthe graph of each equation is a parabola, circle, ellipse, or hyperbola.

a. 2x2 ! y2 $ 6x ! 8y ! 12 # 0 A # ; C # ; type of graph:

b. 2x2 ! 3x $ 2y $ 5 # 0 A # ; C # ; type of graph:

c. 5x2 ! 10x ! 5y2 $ 20y ! 1 # 0 A # ; C # ; type of graph:

d. x2 $ y2 ! 4x ! 2y $ 5 # 0 A # ; C # ; type of graph:

Helping You Remember

3. What is an easy way to recognize that an equation represents a parabola rather thanone of the other conic sections?

If the equation has an x2 term and y term but no y2 term, then the graphis a parabola. Likewise, if the equation has a y2 term and x term but nox2 term, then the graph is a parabola.

hyperbola!11circle55

parabola02ellipse12

( y ! 5)2"15

(x $ 3)2"36

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© Glencoe/McGraw-Hill 490 Glencoe Algebra 2

LociA locus (plural, loci) is the set of all points, and only those points, that satisfya given set of conditions. In geometry, figures often are defined as loci. Forexample, a circle is the locus of points of a plane that are a given distancefrom a given point. The definition leads naturally to an equation whose graphis the curve described.

Write an equation of the locus of points that are thesame distance from (3, 4) and y % !4.

Recognizing that the locus is a parabola with focus (3, 4) and directrix y # $4,you can find that h # 3, k # 0, and a # 4 where (h, k) is the vertex and 4 unitsis the distance from the vertex to both the focus and directrix.

Thus, an equation for the parabola is y # "116"(x $ 3)2.

The problem also may be approached analytically as follows:

Let (x, y) be a point of the locus.

The distance from (3, 4) to (x, y) # the distance from y # $4 to (x, y).

$(x $ 3#)2 ! (#y $ 4)#2# # $(x $ x#)2 ! (#y $ ($#4))2#(x $ 3)2 ! y2 $ 8y ! 16 # y2 ! 8y ! 16

(x $ 3)2 # 16y

"116"(x $ 3)2 # y

Describe each locus as a geometric figure. Then write an equationfor the locus.

1. All points that are the same distance from (0, 5) and (4, 5).

2. All points that are 4 units from the origin.

3. All points that are the same distance from ($2, $1) and x # 2.

4. The locus of points such that the sum of the distances from ($2, 0) and (2, 0) is 6.

5. The locus of points such that the absolute value of the difference of the distances from ($3, 0) and (3, 0) is 2.

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

8-68-6

ExampleExample

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Study Guide and InterventionSolving Quadratic Systems

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

© Glencoe/McGraw-Hill 491 Glencoe Algebra 2

Less

on

8-7

Systems of Quadratic Equations Like systems of linear equations, systems ofquadratic equations can be solved by substitution and elimination. If the graphs are a conicsection and a line, the system will have 0, 1, or 2 solutions. If the graphs are two conicsections, the system will have 0, 1, 2, 3, or 4 solutions.

Solve the system of equations. y % x2 ! 2x ! 15x # y % !3

Rewrite the second equation as y # $x $ 3 and substitute into the first equation.

$x $ 3 # x2 $ 2x $ 150 # x2 $ x $ 12 Add x ! 3 to each side.

0 # (x $ 4)(x ! 3) Factor.

Use the Zero Product property to getx # 4 or x # $3.

Substitute these values for x in x ! y # $3:

4 ! y # $3 or $3 ! y # $3y # $7 y # 0

The solutions are (4, $7) and ($3, 0).

Find the exact solution(s) of each system of equations.

1. y# x2 $ 5 2. x2 ! ( y $ 5)2 # 25y# x $ 3 y # $x2

(2, !1), (!1, !4) (0, 0)

3. x2 ! ( y $ 5)2 # 25 4. x2 ! y2 # 9y # x2 x2 ! y # 3

(0, 0), (3, 9), (!3, 9) (0, 3), ($5!, !2), (!$5!, !2)

5. x2 $ y2 # 1 6. y # x $ 3x2 ! y2 # 16 x # y2 $ 4

" , #, " , ! #, " , #, "! , #, "! , ! # " , #1 ! $29!""2

7 ! $29!""2$30!"$34!"$30!"$34!"

1 # $29!""27 # $29!""2

$30!"$34!"$30!"$34!"

ExampleExample

ExercisesExercises

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© Glencoe/McGraw-Hill 492 Glencoe Algebra 2

Systems of Quadratic Inequalities Systems of quadratic inequalities can be solvedby graphing.

Solve the system of inequalities by graphing.x2 # y2 ) 25

"x ! #2# y2 *

The graph of x2 ! y2 * 25 consists of all points on or inside the circle with center (0, 0) and radius 5.The graph of

!x $ "2! y2 + consists of all points on or outside the

circle with center ! , 0" and radius . The solution of the

system is the set of points in both regions.

Solve the system of inequalities by graphing.x2 # y2 ) 25

! & 1

The graph of x2 ! y2 * 25 consists of all points on or inside the circle with center (0, 0) and radius 5.The graph of

$ & 1 are the points “inside” but not on the branches of

the hyperbola shown. The solution of the system is the set ofpoints in both regions.

Solve each system of inequalities below by graphing.

1. ! * 1 2. x2 ! y2 * 169 3. y + (x $ 2)2

y & x $ 2x2 ! 9y2 + 225 (x ! 1)2 ! ( y ! 1)2 * 16

x

y

Ox

y

O 6 12

12

6

–6

–12

–6–12x

y

O

1"2

y2"4

x2"16

x2"9

y2"4

x2"9

y2"4

x

y

O

5"2

5"2

25"4

5"2

25"4

5"2

x

y

O

Study Guide and Intervention (continued)

Solving Quadratic Systems

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

Example 1Example 1

Example 2Example 2

ExercisesExercises

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Skills PracticeSolving Quadratic Systems

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

© Glencoe/McGraw-Hill 493 Glencoe Algebra 2

Less

on

8-7

Find the exact solution(s) of each system of equations.

1. y # x $ 2 (0, !2), (1, !1) 2. y # x ! 3 (!1, 2), 3. y # 3x (0, 0)y # x2 $ 2 y # 2x2 (1.5, 4.5) x # y2

4. y # x ($2!, $2!), 5. x # $5 (!5, 0) 6. y # 7 no solutionx2 ! y2 # 4 (!$2!, !$2!) x2 ! y2 # 25 x2 ! y2 # 9

7. y # $2x ! 2 (2, !2), 8. x $ y ! 1 # 0 (1, 2) 9. y # 2 $ x (0, 2), (3,!1)y2 # 2x " , 1# y2 # 4x y # x2 $ 4x ! 2

10. y # x $ 1 no solution 11. y # 3x2 (0, 0) 12. y # x2 ! 1 (!1, 2), y # x2 y # $3x2 y # $x2 ! 3 (1, 2)

13. y # 4x (!1, !4), (1, 4) 14. y # $1 (0, !1) 15. 4x2 ! 9y2 # 36 (!3, 0), 4x2 ! y2 # 20 4x2 ! y2 # 1 x2 $ 9y2 # 9 (3, 0)

16. 3( y ! 2)2 $ 4(x $ 3)2 # 12 17. x2 $ 4y2 # 4 (!2, 0), 18. y2 $ 4x2 # 4 no y # $2x ! 2 (0, 2), (3, !4) x2 ! y2 # 4 (2, 0) y # 2x solution

Solve each system of inequalities by graphing.

19. y * 3x $ 2 20. y * x 21. 4y2 ! 9x2 ' 144x2 ! y2 ' 16 y + $2x2 ! 4 x2 ! 8y2 ' 16

x

y

O 4 8

8

4

–4

–8

–4–8x

y

Ox

y

O

1"

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Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
Brad and Anissa Stuckey
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© Glencoe/McGraw-Hill 494 Glencoe Algebra 2

Find the exact solution(s) of each system of equations.

1. (x $ 2)2 ! y2 # 5 2. x # 2( y ! 1)2 $ 6 3. y2 $ 3x2 # 6 4. x2 ! 2y2 # 1x $ y # 1 x ! y # 3 y # 2x $ 1 y # $x ! 1

(0, !1), (3, 2) (2, 1), (6.5, !3.5) (!1, !3), (5, 9) (1, 0), " , #5. 4y2 $ 9x2 # 36 6. y # x2 $ 3 7. x2 ! y2 # 25 8. y2 # 10 $ 6x2

4x2 $ 9y2 # 36 x2 ! y2 # 9 4y # 3x 4y2 # 40 $ 2x2

no solution (0, !3), (($5!, 2) (4, 3), (!4, !3) (0, ($10! )9. x2 ! y2 # 25 10. 4x2 ! 9y2 # 36 11. x # $( y $ 3)2 ! 2 12. $ # 1

x # 3y $ 5 2x2 $ 9y2 # 18 x # ( y $ 3)2 ! 3x2 ! y2 # 9

(!5, 0), (4, 3) ((3, 0) no solution ((3, 0)

13. 25x2 ! 4y2 # 100 14. x2 ! y2 # 4 15. x2 $ y2 # 3

x # $ ! # 1 y2 $ x2 # 3

no solution ((2, 0) no solution

16. ! # 1 17. x ! 2y # 3 18. x2 ! y2 # 64

3x2 $ y2 # 9x2 ! y2 # 9 x2 $ y2 # 8

((2, ($3!) (3, 0), "! , # ((6, (2$7!)Solve each system of inequalities by graphing.

19. y + x2 20. x2 ! y2 ' 36 21. ! * 1y & $x ! 2 x2 ! y2 + 16

(x ! 1)2 ! ( y $ 2)2 * 4

22. GEOMETRY The top of an iron gate is shaped like half an ellipse with two congruent segments from the center of theellipse to the ellipse as shown. Assume that the center ofthe ellipse is at (0, 0). If the ellipse can be modeled by theequation x2 ! 4y2 # 4 for y + 0 and the two congruent

segments can be modeled by y # x and y # $ x,

what are the coordinates of points A and B?

$3#"2

$3#"2

BA

(0, 0)

x

y

O

x

y

O 4 8

8

4

–4

–8

–4–8

x

y

O

(x ! 2)2"4

( y $ 3)2"16

12"

9"

y2"7

x2"7

y2"8

x2"4

5"2

y2"16

x2"9

2"

1"

Practice (Average)

Solving Quadratic Systems

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

"!1, # and "1, #$3!"$3!"

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Reading to Learn MathematicsSolving Quadratic Systems

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

© Glencoe/McGraw-Hill 495 Glencoe Algebra 2

Less

on

8-7

Pre-Activity How do systems of equations apply to video games?

Read the introduction to Lesson 8-7 at the top of page 455 in your textbook.

The figure in your textbook shows that the spaceship hits the circular forcefield in two points. Is it possible for the spaceship to hit the force field ineither fewer or more than two points? State all possibilities and explainhow these could happen. Sample answer: The spaceship could hitthe force field in zero points if the spaceship missed the forcefield all together. The spaceship could also hit the force fieldin one point if the spaceship just touched the edge of theforce field.

Reading the Lesson

1. Draw a sketch to illustrate each of the following possibilities.

a. a parabola and a line b. an ellipse and a circle c. a hyperbola and athat intersect in that intersect in line that intersect in2 points 4 points 1 point

2. Consider the following system of equations.

x2 # 3 ! y2

2x2 ! 3y2 # 11

a. What kind of conic section is the graph of the first equation? hyperbolab. What kind of conic section is the graph of the second equation? ellipsec. Based on your answers to parts a and b, what are the possible numbers of solutions

that this system could have? 0, 1, 2, 3, or 4

Helping You Remember

3. Suppose that the graph of a quadratic inequality is a region whose boundary is a circle.How can you remember whether to shade the interior or the exterior of the circle?Sample answer: The solutions of an inequality of the form x2 # y2 ' r2

are all points that are less than r units from the origin, so the graph isthe interior of the circle. The solutions of an inequality of the form x2 # y2 & r2 are the points that are more than r units from the origin, sothe graph is the exterior of the circle.

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© Glencoe/McGraw-Hill 496 Glencoe Algebra 2

Graphing Quadratic Equations with xy-TermsYou can use a graphing calculator to examine graphs of quadratic equations that contain xy-terms.

Use a graphing calculator to display the graph of x2 # xy # y2 % 4.

Solve the equation for y in terms of x by using the quadratic formula.

y2 ! xy ! (x2 $ 4) # 0

To use the formula, let a # 1, b # x, and c # (x2 $ 4).

y #

y #

To graph the equation on the graphing calculator, enter the two equations:

y # and y #

Use a graphing calculator to graph each equation. State the type of curve each graph represents.

1. y2 ! xy # 8 2. x2 ! y2 $ 2xy $ x # 0

3. x2 $ xy ! y2 # 15 4. x2 ! xy ! y2 # $9

5. 2x2 $ 2xy $ y2 ! 4x # 20 6. x2 $ xy $ 2y2 ! 2x ! 5y $ 3 # 0

$x $ $16 $#3x2#"""2

$x ! $16 $#3x2#"""2

$x ( $16 $#3x2#"""2

$x ( $x2 $ 4#(1)(x2#$ 4)#"""2

x

y

O 1–1–2 2

2

1

–1

–2

Enrichment

NAME ______________________________________________ DATE ____________ PERIOD _____

8-78-7

ExampleExample

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© Glencoe/McGraw-Hill A2 Glencoe Algebra 2

Answers (Lesson 8-1)

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ide

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____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

©G

lenc

oe/M

cGra

w-Hi

ll45

5G

lenc

oe A

lgeb

ra 2

Lesson 8-1

The

Mid

po

int

Form

ula

Mid

poin

t For

mul

aTh

e m

idpo

int M

of a

seg

men

t with

end

poin

ts (x

1, y 1

) and

(x2,

y 2) i

s !

, ".

y 1!

y 2"

2x 1

!x 2

"2

Fin

d t

he

mid

poi

nt

of t

he

lin

e se

gmen

t w

ith

en

dp

oin

ts a

t (4

,!7)

an

d (

!2,

3).

!,

"#!

,"

#!

,"o

r (1

,$2)

The

mid

poin

t of

the

seg

men

t is

(1,

$2)

.

$4

"2

2 " 2

$7

!3

"2

4 !

($2)

"" 2

y 1!

y 2"

2x 1

!x 2

"2

A d

iam

eter

A!B!

of a

cir

cle

has

en

dp

oin

ts A

(5,!

11)

and

B(!

7,6)

.W

hat

are

th

e co

ord

inat

es o

f th

e ce

nte

rof

th

e ci

rcle

?

The

cen

ter

of t

he c

ircl

e is

the

mid

poin

t of

all

of it

s di

amet

ers.

!,

"#!

,"

#!

,"o

r !$

1,$

2"

The

cir

cle

has

cent

er !$

1,$

2".

1 " 2

1 " 2$

5"2

$2

"2

$11

!6

"" 2

5 !

($7)

"" 2

y 1!

y 2"

2x 1

!x 2

"2

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Fin

d t

he

mid

poi

nt

of e

ach

lin

e se

gmen

t w

ith

en

dp

oin

ts a

t th

e gi

ven

coo

rdin

ates

.

1.(1

2,7)

and

($

2,11

)2.

($8,

$3)

and

(10

,9)

3.(4

,15)

and

(10

,1)

(5,9

)(1

,3)

(7,8

)

4.($

3,$

3) a

nd (

3,3)

5.(1

5,6)

and

(12

,14)

6.(2

2,$

8) a

nd (

$10

,6)

(0,0

)(1

3.5,

10)

(6,!

1)

7.(3

,5)

and

($6,

11)

8.(8

,$15

) an

d ($

7,13

)9.

(2.5

,$6.

1) a

nd (

7.9,

13.7

)

"!,8

#"

,!1 #

(5.2

,3.8

)

10.(

$7,

$6)

and

($

1,24

) 11

.(3,

$10

) an

d (3

0,$

20)

12.(

$9,

1.7)

and

($

11,1

.3)

(!4,

9)"

,!15

#(!

10,1

.5)

13.S

egm

ent

M#N#

has

mid

poin

t P

.If

Mha

s co

ordi

nate

s (1

4,$

3) a

nd P

has

coor

dina

tes

($8,

6),w

hat

are

the

coor

dina

tes

of N

?(!

30,1

5)

14.C

ircl

e R

has

a di

amet

er S#

T#.I

f R

has

coor

dina

tes

($4,

$8)

and

Sha

s co

ordi

nate

s (1

,4),

wha

t ar

e th

e co

ordi

nate

s of

T?

(!9,

!20

)

15.S

egm

ent

A#D#

has

mid

poin

t B

,and

B#D#

has

mid

poin

t C

.If A

has

coor

dina

tes

($5,

4) a

nd

Cha

s co

ordi

nate

s (1

0,11

),w

hat

are

the

coor

dina

tes

of B

and

D?

Bis

"5,8

#,Dis

"15,

13#.1 " 3

2 " 3

33 " 21 " 23 " 2

©G

lenc

oe/M

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w-Hi

ll45

6G

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oe A

lgeb

ra 2

The

Dis

tan

ce F

orm

ula

Dist

ance

For

mul

aTh

e di

stan

ce b

etwe

en tw

o po

ints

(x1,

y 1) a

nd (x

2, y 2

) is

give

n by

d

#$

(x2

$#

x 1)2

!#

(y2

$#

y 1)2

#.

Wh

at i

s th

e d

ista

nce

bet

wee

n (

8,!

2) a

nd

(!

6,!

8)?

d#

$(x

2$

#x 1

)2!

#(y

2$

#y 1

)2#

Dist

ance

For

mul

a

#$

($6

$#

8)2

!#

[$8

$#

($2)

]#

2 #Le

t (x 1

, y1)

#(8

, $2)

and

(x2,

y 2) #

($6,

$8)

.

#$

($14

)#

2!

($#

6)2

#Su

btra

ct.

#$

196

!#

36#or

$23

2#

Sim

plify

.

The

dis

tanc

e be

twee

n th

e po

ints

is $

232

#or

abo

ut 1

5.2

unit

s.

Fin

d t

he

per

imet

er a

nd

are

a of

squ

are

PQ

RS

wit

h v

erti

ces

P(!

4,1)

,Q

(!2,

7),R

(4,5

),an

d S

(2,!

1).

Fin

d th

e le

ngth

of

one

side

to

find

the

per

imet

er a

nd t

he a

rea.

Cho

ose

P#Q#.

d#

$(x

2$

#x 1

)2!

#(y

2$

#y 1

)2#

Dist

ance

For

mul

a

#$

[$4

$#

($2)

]#

2!

(1#

$ 7

)2#

Let (

x 1, y

1) #

($4,

1) a

nd (x

2, y 2

) #($

2, 7

).

#$

($2)

2#

!($

6#

)2 #Su

btra

ct.

#$

40#or

2$

10#Si

mpl

ify.

Sinc

e on

e si

de o

f th

e sq

uare

is 2

$10#

,the

per

imet

er is

8$

10#un

its.

The

are

a is

(2$

10#)2 ,o

r40

uni

ts2 .

Fin

d t

he

dis

tan

ce b

etw

een

eac

h p

air

of p

oin

ts w

ith

th

e gi

ven

coo

rdin

ates

.

1.(3

,7)

and

($1,

4)

2.($

2,$

10)

and

(10,

$5)

3.

(6,$

6) a

nd (

$2,

0)

5 un

its13

uni

ts10

uni

ts4.

(7,2

) an

d (4

,$1)

5.

($5,

$2)

and

(3,

4)

6.(1

1,5)

and

(16

,9)

3$2!

units

10 u

nits

$41!

units

7.($

3,4)

and

(6,

$11

) 8.

(13,

9) a

nd (

11,1

5)

9.($

15,$

7) a

nd (

2,12

)

3$34!

units

2$10!

units

5$26!

units

10.R

ecta

ngle

AB

CD

has

vert

ices

A(1

,4),

B(3

,1),

C($

3,$

2),a

nd D

($5,

1).F

ind

the

peri

met

er a

nd a

rea

of A

BC

D.

2 $13 !

#6 $

5 !un

its;3

$65 !

units

2

11.C

ircl

e R

has

diam

eter

S#T#

wit

h en

dpoi

nts

S(4

,5)

and

T($

2,$

3).W

hat

are

the

circ

umfe

renc

e an

d ar

ea o

f th

e ci

rcle

? (E

xpre

ss y

our

answ

er in

ter

ms

of %

.)10

$un

its;2

5$un

its2

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Mid

poin

t and

Dis

tanc

e Fo

rmul

as

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Page 51: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A3 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-1)

Skil

ls P

ract

ice

Mid

poin

t and

Dis

tanc

e Fo

rmul

as

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

©G

lenc

oe/M

cGra

w-Hi

ll45

7G

lenc

oe A

lgeb

ra 2

Lesson 8-1

Fin

d t

he

mid

poi

nt

of e

ach

lin

e se

gmen

t w

ith

en

dp

oin

ts a

t th

e gi

ven

coo

rdin

ates

.

1.(4

,$1)

,($

4,1)

(0,0

)2.

($1,

4),(

5,2)

(2,3

)

3.(3

,4),

(5,4

)(4

,4)

4.(6

,2),

(2,$

1)"4,

#

5.(3

,9),

($2,

$3)

",3

#6.

($3,

5),(

$3,

$8)

"!3,

!#

7.(3

,2),

($5,

0)(!

1,1)

8.(3

,$4)

,(5,

2)(4

,!1)

9.($

5,$

9),(

5,4)

"0,!

#10

.($

11,1

4),(

0,4)

"!,9

#

11.(

3,$

6),(

$8,

$3)

"!,!

#12

.(0,

10),

($2,

$5)

"!1,

#

Fin

d t

he

dis

tan

ce b

etw

een

eac

h p

air

of p

oin

ts w

ith

th

e gi

ven

coo

rdin

ates

.

13.(

4,12

),($

1,0)

13 u

nits

14.(

7,7)

,($

5,$

2)15

uni

ts

15.(

$1,

4),(

1,4)

2 un

its16

.(11

,11)

,(8,

15)

5 un

its

17.(

1,$

6),(

7,2)

10 u

nits

18.(

3,$

5),(

3,4)

9 un

its

19.(

2,3)

,(3,

5)$

5!un

its20

.($

4,3)

,($

1,7)

5 un

its

21.(

$5,

$5)

,(3,

10)

17 u

nits

22.(

3,9)

,($

2,$

3)13

uni

ts

23.(

6,$

2),(

$1,

3)$

74!un

its24

.($

4,1)

,(2,

$4)

$61!

units

25.(

0,$

3),(

4,1)

4$2!

units

26.(

$5,

$6)

,(2,

0)$

85!un

its

5 " 29 " 2

5 " 2

11 " 25 " 2

3 " 21 " 2

1 " 2

©G

lenc

oe/M

cGra

w-Hi

ll45

8G

lenc

oe A

lgeb

ra 2

Fin

d t

he

mid

poi

nt

of e

ach

lin

e se

gmen

t w

ith

en

dp

oin

ts a

t th

e gi

ven

coo

rdin

ates

.

1.(8

,$3)

,($

6,$

11)

(1,!

7)2.

($14

,5),

(10,

6)"!

2,#

3.($

7,$

6),(

1,$

2)(!

3,!

4)4.

(8,$

2),(

8,$

8)(8

,!5)

5.(9

,$4)

,(1,

$1)

"5,!

#6.

(3,3

),(4

,9)"

,6#

7.(4

,$2)

,(3,

$7)

",!

#8.

(6,7

),(4

,4)"5,

#9.

($4,

$2)

,($

8,2)

(!6,

0)10

.(5,

$2)

,(3,

7)"4,

#11

.($

6,3)

,($

5,$

7)"!

,!2 #

12.(

$9,

$8)

,(8,

3)"!

,!#

13.(

2.6,

$4.

7),(

8.4,

2.5)

(5.5

,!1.

1)14

. !$,6

", !,4

" ",5

#15

.($

2.5,

$4.

2),(

8.1,

4.2)

(2.8

,0)

16. !

,", !

$,$

" "!

,0#

Fin

d t

he

dis

tan

ce b

etw

een

eac

h p

air

of p

oin

ts w

ith

th

e gi

ven

coo

rdin

ates

.

17.(

5,2)

,(2,

$2)

5 un

its18

.($

2,$

4),(

4,4)

10 u

nits

19.(

$3,

8),(

$1,

$5)

$17

3!

units

20.(

0,1)

,(9,

$6)

$13

0!

units

21.(

$5,

6),(

$6,

6)1

unit

22.(

$3,

5),(

12,$

3)17

uni

ts

23.(

$2,

$3)

,(9,

3)$

157

!un

its24

.($

9,$

8),(

$7,

8)2$

65!un

its

25.(

9,3)

,(9,

$2)

5 un

its26

.($

1,$

7),(

0,6)

$17

0!

units

27.(

10,$

3),(

$2,

$8)

13 u

nits

28.(

$0.

5,$

6),(

1.5,

0)2$

10!un

its

29. !

,", !

1,"1

uni

t30

.($

4$2#,

$$

5#),(

$5$

2#,4$

5#)$

127

!un

its

31.G

EOM

ETRY

Cir

cle

Oha

s a

diam

eter

A#B#

.If

Ais

at

($6,

$2)

and

Bis

at

($3,

4),f

ind

the

cent

er o

f th

e ci

rcle

and

the

leng

th o

f it

s di

amet

er."!

,1#;3

$5!

units

32.G

EOM

ETRY

Fin

d th

e pe

rim

eter

of a

tri

angl

e w

ith

vert

ices

at

(1,$

3),(

$4,

9),a

nd ($

2,1)

.18

#2$

17!un

its

9 " 2

7 " 53 " 5

2 " 5

1 " 41 " 2

5 " 81 " 2

1 " 8

1 " 62 " 3

1 " 3

5 " 21 " 2

11 " 2

5 " 2

11 " 29 " 2

7 " 2

7 " 25 " 2

11 " 2

Pra

ctic

e (A

vera

ge)

Mid

poin

t and

Dis

tanc

e Fo

rmul

as

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

Page 52: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A4 Glencoe Algebra 2

Answers (Lesson 8-1)

Rea

din

g t

o L

earn

Math

emati

csM

idpo

int a

nd D

ista

nce

Form

ulas

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

©G

lenc

oe/M

cGra

w-Hi

ll45

9G

lenc

oe A

lgeb

ra 2

Lesson 8-1

Pre-

Act

ivit

yH

ow a

re t

he

Mid

poi

nt

and

Dis

tan

ce F

orm

ula

s u

sed

in

em

erge

ncy

med

icin

e?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

1 at

the

top

of

page

412

in y

our

text

book

.

How

do

you

find

dis

tanc

es o

n a

road

map

?

Sam

ple

answ

er:U

se th

e sc

ale

of m

iles

on th

e m

ap.Y

ou m

ight

also

use

a ru

ler.

Rea

din

g t

he

Less

on

1.a.

Wri

te t

he c

oord

inat

es o

f th

e m

idpo

int

of a

seg

men

t w

ith

endp

oint

s (x

1,y 1

) an

d (x

2,y 2

).

",

#b.

Exp

lain

how

to

find

the

mid

poin

t of

a s

egm

ent

if y

ou k

now

the

coo

rdin

ates

of

the

endp

oint

s.D

o no

t us

e su

bscr

ipts

in y

our

expl

anat

ion.

Sam

ple

answ

er:T

o fin

d th

e x-

coor

dina

te o

f the

mid

poin

t,ad

d th

e x-

coor

dina

tes

of th

e en

dpoi

nts

and

divi

de b

y tw

o.To

find

the

y-co

ordi

nate

of t

he m

idpo

int,

do th

e sa

me

with

the

y-co

ordi

nate

s of

the

endp

oint

s.

2.a.

Wri

te a

n ex

pres

sion

for

the

dis

tanc

e be

twee

n tw

o po

ints

wit

h co

ordi

nate

s (x

1,y 1

) an

d(x

2,y 2

).$

(x2

!!

x 1)2

#!

(y2

!!

y 1)2

!b.

Exp

lain

how

to

find

the

dis

tanc

e be

twee

n tw

o po

ints

.Do

not

use

subs

crip

ts in

you

rex

plan

atio

n.

Sam

ple

answ

er:F

ind

the

diffe

renc

e be

twee

n th

e x-

coor

dina

tes

and

squa

re it

.Fin

d th

e di

ffere

nce

betw

een

the

y-co

ordi

nate

s an

d sq

uare

it.A

dd th

e sq

uare

s.Th

en fi

nd th

e sq

uare

root

of t

he s

um.

3.C

onsi

der

the

segm

ent

conn

ecti

ng t

he p

oint

s ($

3,5)

and

(9,

11).

a.F

ind

the

mid

poin

t of

thi

s se

gmen

t.(3

,8)

b.F

ind

the

leng

th o

f th

e se

gmen

t.W

rite

you

r an

swer

in s

impl

ifie

d ra

dica

l for

m.

6$5!

Hel

pin

g Y

ou

Rem

emb

er

4.H

ow c

an t

he “

mid

”in

mid

poin

t he

lp y

ou r

emem

ber

the

mid

poin

t fo

rmul

a?

Sam

ple

answ

er:T

he m

idpo

inti

s th

e po

int i

n th

e m

iddl

eof

a s

egm

ent.

Itis

hal

fway

bet

wee

n th

e en

dpoi

nts.

The

coor

dina

tes

of th

e m

idpo

int a

refo

und

by fi

ndin

g th

e av

erag

e of

the

two

x-co

ordi

nate

s (a

dd th

em a

nddi

vide

by

2) a

nd th

e av

erag

e of

the

two

y-co

ordi

nate

s.

y 1#

y 2"

2x 1

#x 2

"2

©G

lenc

oe/M

cGra

w-Hi

ll46

0G

lenc

oe A

lgeb

ra 2

Qua

drat

ic F

orm

Con

side

r tw

o m

etho

ds f

or s

olvi

ng t

he f

ollo

win

g eq

uati

on.

(y$

2)2

$5(

y$

2) !

6#

0

One

way

to

solv

e th

e eq

uati

on is

to

sim

plif

y fi

rst,

then

use

fac

tori

ng.

y2$

4y!

4 $

5y!

10 !

6#

0y2

$9y

!20

#0

(y$

4)(y

$5)

#0

Thu

s,th

e so

luti

on s

et is

{4,

5}.

Ano

ther

way

to

solv

e th

e eq

uati

on is

fir

st t

o re

plac

e y

$2

by a

sin

gle

vari

able

.T

his

will

pro

duce

an

equa

tion

tha

t is

eas

ier

to s

olve

tha

n th

e or

igin

al e

quat

ion.

Let

t#

y$

2 an

d th

en s

olve

the

new

equ

atio

n.

(y$

2)2

$5(

y$

2) !

6#

0t2

$5t

!6

#0

(t$

2)(t

$3)

#0

Thu

s,t

is 2

or

3.Si

nce

t#

y$

2,th

e so

luti

on s

et o

f th

e or

igin

al e

quat

ion

is {

4,5}

.

Sol

ve e

ach

equ

atio

n u

sin

g tw

o d

iffe

ren

t m

eth

ods.

1.(z

!2)

2!

8(z

!2)

!7

#0

2.(3

x$

1)2

$(3

x$

1) $

20 #

0

{!3,

!9}

{2,!

1}

3.(2

t!

1)2

$4(

2t!

1) !

3 #

04.

(y2

$1)

2$

(y2

$1)

$2

#0

{0,1

}&0,

($

3!'

5.(a

2$

2)2

$2(

a2$

2) $

3 #

06.

(1 !

$c#)

2!

(1 !

$c#)

$6

#0

&(1,

($

5!'{1

}

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-1

8-1

Page 53: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A5 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-2)

Stu

dy

Gu

ide

and I

nte

rven

tion

Para

bola

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

©G

lenc

oe/M

cGra

w-Hi

ll46

1G

lenc

oe A

lgeb

ra 2

Lesson 8-2

Equ

atio

ns

of

Para

bo

las

A p

arab

ola

is a

cur

ve c

onsi

stin

g of

all

poin

ts in

the

coor

dina

te p

lane

tha

t ar

e th

e sa

me

dist

ance

fro

m a

giv

en p

oint

(th

e fo

cus)

and

a g

iven

line

(the

dir

ectr

ix).

The

fol

low

ing

char

t su

mm

ariz

es im

port

ant

info

rmat

ion

abou

t pa

rabo

las.

Stan

dard

For

m o

f Equ

atio

ny

#a(

x$

h)2

!k

x#

a(y

$k)

2!

h

Axis

of S

ymm

etry

x#

hy

#k

Vert

ex(h

, k)

(h, k

)

Focu

s!h,

k!

"!h

!, k

"Di

rect

rixy

#k

$x

#h

$

Dire

ctio

n of

Ope

ning

upwa

rd if

a&

0, d

ownw

ard

if a

'0

right

if a

&0,

left

if a

'0

Leng

th o

f Lat

us R

ectu

m

uni

ts

uni

ts

Iden

tify

th

e co

ord

inat

es o

f th

e ve

rtex

an

d f

ocu

s,th

e eq

uat

ion

s of

the

axis

of

sym

met

ry a

nd

dir

ectr

ix,a

nd

th

e d

irec

tion

of

open

ing

of t

he

par

abol

aw

ith

equ

atio

n y

%2x

2!

12x

!25

.

y#

2x2

$12

x$

25O

rigin

al e

quat

ion

y#

2(x2

$6x

) $25

Fact

or 2

from

the

x-te

rms.

y#

2(x2

$6x

!■

) $25

$2(■

)Co

mpl

ete

the

squa

re o

n th

e rig

ht s

ide.

y#

2(x2

$6x

!9)

$25

$2(

9)Th

e 9

adde

d to

com

plet

e th

e sq

uare

is m

ultip

lied

by 2

.y

#2(

x$

3)2

$43

Writ

e in

sta

ndar

d fo

rm.

The

ver

tex

of t

his

para

bola

is lo

cate

d at

(3,

$43

),th

e fo

cus

is lo

cate

d at

!3,$

42",t

he

equa

tion

of

the

axis

of

sym

met

ry is

x#

3,an

d th

e eq

uati

on o

f th

e di

rect

rix

is y

#$

43.

The

par

abol

a op

ens

upw

ard.

Iden

tify

th

e co

ord

inat

es o

f th

e ve

rtex

an

d f

ocu

s,th

e eq

uat

ion

s of

th

e ax

is o

fsy

mm

etry

an

d d

irec

trix

,an

d t

he

dir

ecti

on o

f op

enin

g of

th

e p

arab

ola

wit

h t

he

give

n e

quat

ion

.

1.y

#x2

!6x

$4

2.y

#8x

$2x

2!

103.

x#

y2$

8y!

6

(!3,

!13

),(2

,18)

, "2,1

7#,

(!10

,4),

"!9

,4#,

"!3,

!12

#,x%

!3,

x%

2,y

%18

,y

%4,

x%

!10

,

y%

!13

,up

dow

nrig

ht

Wri

te a

n e

quat

ion

of

each

par

abol

a d

escr

ibed

bel

ow.

4.fo

cus

($2,

3),d

irec

trix

x#

$2

5.ve

rtex

(5,

1),f

ocus

!4,1

"x

%6(

y!

3)2

!2

x%

!3(

y!

1)2

#5

1 " 24

11 " 121 " 12

1 " 4

1 " 41 " 8

3 " 4

3 " 41 " 8

1 " 8

7 " 8

1 " a1 " a

1 " 4a1 " 4a

1 " 4a1 " 4a

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll46

2G

lenc

oe A

lgeb

ra 2

Gra

ph

Par

abo

las

To g

raph

an

equa

tion

for

a p

arab

ola,

firs

t pu

t th

e gi

ven

equa

tion

inst

anda

rd f

orm

.

y#

a(x

$h)

2!

kfo

r a

para

bola

ope

ning

up

or d

own,

orx

#a(

y$

k)2

!h

for

a pa

rabo

la o

peni

ng t

o th

e le

ft o

r ri

ght

Use

the

val

ues

of a

,h,a

nd k

to d

eter

min

e th

e ve

rtex

,foc

us,a

xis

of s

ymm

etry

,and

leng

th o

fth

e la

tus

rect

um.T

he v

erte

x an

d th

e en

dpoi

nts

of t

he la

tus

rect

um g

ive

thre

e po

ints

on

the

para

bola

.If

you

need

mor

e po

ints

to

plot

an

accu

rate

gra

ph,s

ubst

itut

e va

lues

for

poi

nts

near

the

ver

tex.

Gra

ph

y%

(x!

1)2

#2.

In t

he e

quat

ion,

a#

,h#

1,k

#2.

The

par

abol

a op

ens

up,s

ince

a&

0.ve

rtex

:(1,

2)ax

is o

f sy

mm

etry

:x#

1

focu

s:!1,

2 !

"or !1

,2"

leng

th o

f la

tus

rect

um:

or 3

uni

ts

endp

oint

s of

latu

s re

ctum

: !2,2

", !$

,2"

Th

e co

ord

inat

es o

f th

e fo

cus

and

th

e eq

uat

ion

of

the

dir

ectr

ix o

f a

par

abol

a ar

egi

ven

.Wri

te a

n e

quat

ion

for

eac

h p

arab

ola

and

dra

w i

ts g

rap

h.

1.(3

,5),

y#

12.

(4,$

4),y

#$

63.

(5,$

1),x

#3

y%

(x!

3)2

#3

y%

(x!

4)2

!5

x%

(y#

1)2

#4

1 " 41 " 4

1 " 8

x

y

Ox

y

O

x

y

O

3 " 41 " 2

3 " 41 " 2

1 " "1 3"

3 " 41

" 4 !"1 3" "x

y

O

1 " 3

1 " 3

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Para

bola

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

Page 54: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A6 Glencoe Algebra 2

Answers (Lesson 8-2)

Skil

ls P

ract

ice

Para

bola

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

©G

lenc

oe/M

cGra

w-Hi

ll46

3G

lenc

oe A

lgeb

ra 2

Lesson 8-2

Wri

te e

ach

equ

atio

n i

n s

tan

dar

d f

orm

.

1.y

#x2

!2x

!2

2.y

#x2

$2x

!4

3.y

#x2

!4x

!1

y%

[x!

(!1)

]2#

1y

%(x

!1)

2#

3y

%[x

!(!

2)]2

#(!

3)

Iden

tify

th

e co

ord

inat

es o

f th

e ve

rtex

an

d f

ocu

s,th

e eq

uat

ion

s of

th

e ax

is o

fsy

mm

etry

an

d d

irec

trix

,an

d t

he

dir

ecti

on o

f op

enin

g of

th

e p

arab

ola

wit

h t

he

give

n e

quat

ion

.Th

en f

ind

th

e le

ngt

h o

f th

e la

tus

rect

um

an

d g

rap

h t

he

par

abol

a.

4.y

#(x

$2)

25.

x#

(y$

2)2

!3

6.y

#$

(x!

3)2

!4

vert

ex:(

2,0)

;ve

rtex

:(3,

2);

vert

ex:(

!3,

4);

focu

s:"2,

#;fo

cus:

"3,2

#;fo

cus:

"!3,3

#;ax

is o

f sym

met

ry:

axis

of s

ymm

etry

:ax

is o

f sym

met

ry:

x%

2;y

%2;

x%

!3;

dire

ctrix

:y%

!;

dire

ctrix

:x%

2;

dire

ctrix

:y%

4;

open

s up

;op

ens

right

;op

ens

dow

n;la

tus

rect

um:1

uni

tla

tus

rect

um:1

uni

tla

tus

rect

um:1

uni

t

Wri

te a

n e

quat

ion

for

eac

h p

arab

ola

des

crib

ed b

elow

.Th

en d

raw

th

e gr

aph

.

7.ve

rtex

(0,

0),

8.ve

rtex

(5,

1),

9.ve

rtex

(1,

3),

focu

s !0,

$"

focu

s !5,

"di

rect

rix

x#

y%

!3x

2y

%(x

!5)

2#

1x

%2(

y!

3)2

#1 x

y

Ox

y

O

x

y

O

7 " 85 " 4

1 " 12

1 " 43 " 4

1 " 4

3 " 41 " 4

1 " 4

x

y

Ox

y

O

x

y

O

©G

lenc

oe/M

cGra

w-Hi

ll46

4G

lenc

oe A

lgeb

ra 2

Wri

te e

ach

equ

atio

n i

n s

tan

dar

d f

orm

.

1.y

#2x

2$

12x

!19

2.y

#x2

!3x

!3.

y#

$3x

2$

12x

$7

y%

2(x

!3)

2#

1y

%[x

!(!

3)]2

#(!

4)y

%!

3[x

!(!

2)]2

#5

Iden

tify

th

e co

ord

inat

es o

f th

e ve

rtex

an

d f

ocu

s,th

e eq

uat

ion

s of

th

e ax

is o

fsy

mm

etry

an

d d

irec

trix

,an

d t

he

dir

ecti

on o

f op

enin

g of

th

e p

arab

ola

wit

h t

he

give

n e

quat

ion

.Th

en f

ind

th

e le

ngt

h o

f th

e la

tus

rect

um

an

d g

rap

h t

he

par

abol

a.

4.y

#(x

$4)

2!

35.

x#

$y2

!1

6.x

#3(

y!

1)2

$3

vert

ex:(

4,3)

;ve

rtex

:(1,

0);

vert

ex:(

!3,

!1)

;fo

cus:

"4,3

#;fo

cus:

",0

#;fo

cus:

"!2

,!1 #;

axis

:x%

4;ax

is:y

%0;

axis

:y%

!1;

dire

ctrix

:y%

2;

dire

ctrix

:x%

1;

dire

ctrix

:x%

!3

;op

ens

up;

open

s le

ft;op

ens

right

;la

tus

rect

um:1

uni

tla

tus

rect

um:3

uni

tsla

tus

rect

um:

unit

Wri

te a

n e

quat

ion

for

eac

h p

arab

ola

des

crib

ed b

elow

.Th

en d

raw

th

e gr

aph

.

7.ve

rtex

(0,

$4)

,8.

vert

ex ($

2,1)

,9.

vert

ex (

1,3)

,

focu

s !0,

$3

"di

rect

rix

x#

$3

axis

of

sym

met

ry x

#1,

latu

s re

ctum

:2 u

nits

,a

'0

y%

2x2

!4

x %

(y !

1)2

!2

y %

!(x

!1)

2#

3

10.T

ELEV

ISIO

NW

rite

the

equ

atio

n in

the

form

y#

ax2

for

a sa

telli

te d

ish.

Ass

ume

that

the

bott

om o

f th

e up

war

d-fa

cing

dis

h pa

sses

thr

ough

(0,

0) a

nd t

hat

the

dist

ance

fro

m t

hebo

ttom

to

the

focu

s po

int

is 8

inch

es.

y%

x21 " 32

x

y

Ox

y

Ox

y

O

1 " 21 " 4

7 " 8

1 " 3

1 " 123 " 4

3 " 4

11 " 121 " 4

1 " 4

x

y

Ox

y

O

x

y

O

1 " 3

1 " 2

1 " 21 " 2

Pra

ctic

e (A

vera

ge)

Para

bola

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

Page 55: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A7 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-2)

Rea

din

g t

o L

earn

Math

emati

csPa

rabo

las

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

©G

lenc

oe/M

cGra

w-Hi

ll46

5G

lenc

oe A

lgeb

ra 2

Lesson 8-2

Pre-

Act

ivit

yH

ow a

re p

arab

olas

use

d i

n m

anu

fact

uri

ng?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

2 at

the

top

of

page

419

in y

our

text

book

.

Nam

e at

leas

t tw

o re

flec

tive

obj

ects

tha

t m

ight

hav

e th

e sh

ape

of a

para

bola

.

Sam

ple

answ

er:t

eles

cope

mirr

or,s

atel

lite

dish

Rea

din

g t

he

Less

on

1.In

the

par

abol

a sh

own

in t

he g

raph

,the

poi

nt (

2,$

2) is

cal

led

the

and

the

poin

t (2

,0)

is c

alle

d th

e

.The

line

y#

$4

is c

alle

d th

e

,and

the

line

x#

2 is

cal

led

the

.

2.a.

Wri

te t

he s

tand

ard

form

of

the

equa

tion

of

a pa

rabo

la t

hat

open

s up

war

d or

dow

nwar

d.y

%a(

x!

h)2

#k

b.T

he p

arab

ola

open

s do

wnw

ard

if

and

open

s up

war

d if

.T

he

equa

tion

of

the

axis

of

sym

met

ry is

,a

nd t

he c

oord

inat

es o

f th

e ve

rtex

are

.

3.A

par

abol

a ha

s eq

uati

on x

#$

(y$

2)2

!4.

Thi

s pa

rabo

la o

pens

to

the

.

It h

as v

erte

x an

d fo

cus

.The

dir

ectr

ix is

.T

he le

ngth

of t

he la

tus

rect

um is

un

its.

Hel

pin

g Y

ou

Rem

emb

er

4.H

ow c

an t

he w

ay in

whi

ch y

ou p

lot

poin

ts in

a r

ecta

ngul

ar c

oord

inat

e sy

stem

hel

p yo

u to

rem

embe

r w

hat

the

sign

of a

tells

you

abo

ut t

he d

irec

tion

in w

hich

a p

arab

ola

open

s?Sa

mpl

e an

swer

:In

plot

ting

poin

ts,a

pos

itive

x-c

oord

inat

e te

lls y

ou to

mov

e to

the

right

and

a ne

gativ

e x-

coor

dina

te te

lls y

ou to

mov

e to

the

left.

This

is li

ke a

par

abol

a w

hose

equ

atio

n is

of t

he fo

rm “x

%…

”;it

open

s to

the

right

if a

&0

and

to th

e le

ftif

a'

0.Li

kew

ise,

a po

sitiv

e y-

coor

dina

te te

lls y

ou to

mov

e up

and

a ne

gativ

e y-

coor

dina

te te

lls y

outo

mov

e do

wn.

This

is li

ke a

par

abol

a w

hose

equ

atio

n is

of t

he fo

rm

“y%

…”;

it op

ens

upw

ard

if a

&0

and

dow

nwar

dif

a'

0.

8x

%6

(2,2

)(4

,2)

left

1 " 8

(h,k

)x

%h

a&

0a

'0

axis

of s

ymm

etry

dire

ctrix

focu

svert

exx

y O

( 2, –

2)

( 2, 0

)

y % –

4

©G

lenc

oe/M

cGra

w-Hi

ll46

6G

lenc

oe A

lgeb

ra 2

Tang

ents

to P

arab

olas

A li

ne t

hat

inte

rsec

ts a

par

abol

a in

exa

ctly

one

poi

nt

wit

hout

cro

ssin

g th

e cu

rve

is a

tan

gen

tto

the

pa

rabo

la.T

he p

oint

whe

re a

tan

gent

line

tou

ches

a

para

bola

is t

he p

oin

t of

tan

gen

cy.T

he li

ne

perp

endi

cula

r to

a t

ange

nt t

o a

para

bola

at

the

poin

t of

tan

genc

yis

cal

led

the

nor

mal

to t

he p

arab

ola

at

that

poi

nt.I

n th

e di

agra

m,l

ine

!is

tan

gent

to

the

para

bola

tha

t is

the

gra

ph o

f y#

x2at

!"3 2" ,"9 4" ".

The

x-ax

is is

tan

gent

to

the

para

bola

at

O,a

nd t

he y

-axi

s is

the

nor

mal

to

the

para

bola

at

O.

Sol

ve e

ach

pro

blem

.

1.F

ind

an e

quat

ion

for

line

!in

the

dia

gram

.Hin

t:A

non

vert

ical

line

wit

h an

equa

tion

of

the

form

y#

mx

!b

will

be

tang

ent

to t

he g

raph

of y

#x2

at

!"3 2" ,"9 4" "i

f an

d on

ly if

!"3 2" ,"9 4" "i

s th

e on

ly p

air

of n

umbe

rs t

hat

sati

sfie

s bo

th

y#

x2an

d y

#m

x!

b.

m%

3,b

%!

"9 4" ,y

%3x

!"9 4"

2.If

ais

any

rea

l num

ber,

then

(a,a

2 ) b

elon

gs t

o th

e gr

aph

of y

#x2

.Exp

ress

m

and

bin

ter

ms

of a

to f

ind

an e

quat

ion

of t

he f

orm

y#

mx

!b

for

the

line

that

is t

ange

nt t

o th

e gr

aph

of y

#x2

at (

a,a2

).

m%

2a,b

%a2

,y%

(2a)

x#

(!a2

)or y

%2a

x!

a2

3.F

ind

an e

quat

ion

for

the

norm

al t

o th

e gr

aph

of y

#x2

at !"3 2" ,

"9 4" ".y

%!

"1 3" x#

"1 41 "

4.If

ais

a n

onze

ro r

eal n

umbe

r,fi

nd a

n eq

uati

on f

or t

he n

orm

al t

o th

e gr

aph

ofy

#x2

at (

a,a2

).

y%

"!" 21 a"

#x#

"a2#

"1 2" #

x

y

O

!

y %

x2

1–1

–2–3

2

6 5 4 3 2 1

3

!3 – 2, 9 – 4"

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-2

8-2

Page 56: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A8 Glencoe Algebra 2

Answers (Lesson 8-3)

Stu

dy

Gu

ide

and I

nte

rven

tion

Circ

les

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-3

8-3

©G

lenc

oe/M

cGra

w-Hi

ll46

7G

lenc

oe A

lgeb

ra 2

Lesson 8-3

Equ

atio

ns

of

Cir

cles

The

equ

atio

n of

a c

ircl

e w

ith

cent

er (h

,k)

and

radi

us r

unit

s is

(x

$h)

2!

(y$

k)2

#r2

.

Wri

te a

n e

quat

ion

for

a c

ircl

e if

th

e en

dp

oin

ts o

f a

dia

met

er a

re a

t(!

4,5)

an

d (

6,!

3).

Use

the

mid

poin

t fo

rmul

a to

fin

d th

e ce

nter

of

the

circ

le.

(h,k

) #!

,"

Mid

poin

t for

mul

a

#!

,"

(x1,

y 1) #

($4,

5),

(x2,

y 2) #

(6, $

3)

#!

,"o

r (1

,1)

Sim

plify

.

Use

the

coo

rdin

ates

of

the

cent

er a

nd o

ne e

ndpo

int

of t

he d

iam

eter

to

find

the

rad

ius.

r#

$(x

2$

x#

1)2

!#

(y2

$#

y 1)2

#Di

stan

ce fo

rmul

a

r#

$($

4 $

#1)

2!

#(5

$#

1)2

#(x

1, y 1

) #(1

, 1),

(x2,

y 2) #

($4,

5)

#$

($5)

2#

!42

##

$41#

Sim

plify

.

The

rad

ius

of t

he c

ircl

e is

$41#

,so

r2#

41.

An

equa

tion

of

the

circ

le is

(x

$1)

2!

(y$

1)2

#41

.

Wri

te a

n e

quat

ion

for

th

e ci

rcle

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

1.ce

nter

(8,

$3)

,rad

ius

6(x

!8)

2#

(y#

3)2

%36

2.ce

nter

(5,

$6)

,rad

ius

4(x

!5)

2#

(y#

6)2

%16

3.ce

nter

($5,

2),p

asse

s th

roug

h ($

9,6)

(x#

5)2

#(y

!2)

2%

32

4.en

dpoi

nts

of a

dia

met

er a

t (6

,6)

and

(10,

12)

(x!

8)2

#(y

!9)

2%

13

5.ce

nter

(3,

6),t

ange

nt t

o th

e x-

axis

(x!

3)2

#(y

!6)

2%

36

6.ce

nter

($4,

$7)

,tan

gent

to

x#

2(x

#4)

2#

(y#

7)2

%36

7.ce

nter

at

($2,

8),t

ange

nt t

o y

#$

4(x

#2)

2#

(y !

8)2

%14

4

8.ce

nter

(7,

7),p

asse

s th

roug

h (1

2,9)

(x!

7)2

#(y

!7)

2%

29

9.en

dpoi

nts

of a

dia

met

er a

re ($

4,$

2) a

nd (

8,4)

(x!

2)2

#(y

!1)

2%

45

10.e

ndpo

ints

of

a di

amet

er a

re ($

4,3)

and

(6,

$8)

(x!

1)2

#(y

#2.

5)2

%55

.25

2 " 22 " 2

5 !

($3)

"" 2

$4

!6

"2

y 1!

y 2"

2x 1

!x 2

"2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll46

8G

lenc

oe A

lgeb

ra 2

Gra

ph

Cir

cles

To g

raph

a c

ircl

e,w

rite

the

giv

en e

quat

ion

in t

he s

tand

ard

form

of

the

equa

tion

of

a ci

rcle

,(x

$h)

2!

(y$

k)2

#r2

.

Plo

t th

e ce

nter

(h,k

) of

the

cir

cle.

The

n us

e r

to c

alcu

late

and

plo

t th

e fo

ur p

oint

s (h

!r,

k),

(h$

r,k)

,(h,

k!

r),a

nd (

h,k

$r)

,whi

ch a

re a

ll po

ints

on

the

circ

le.S

ketc

h th

e ci

rcle

tha

tgo

es t

hrou

gh t

hose

fou

r po

ints

.

Fin

d t

he

cen

ter

and

rad

ius

of t

he

circ

le

wh

ose

equ

atio

n i

s x2

#2x

#y2

#4y

%11

.Th

en g

rap

h

the

circ

le. x2

!2x

!y2

!4y

#11

x2!

2x!

■!

y2!

4y!

■#

11 !

x2!

2x!

1 !

y2!

4y!

4 #

11 !

1 !

4(x

!1)

2!

(y!

2)2

#16

The

refo

re,t

he c

ircl

e ha

s it

s ce

nter

at

($1,

$2)

and

a r

adiu

s of

$

16##

4.Fo

ur p

oint

s on

the

cir

cle

are

(3,$

2),(

$5,

$2)

,($

1,2)

,an

d ($

1,$

6).

Fin

d t

he

cen

ter

and

rad

ius

of t

he

circ

le w

ith

th

e gi

ven

equ

atio

n.T

hen

gra

ph

th

eci

rcle

.

1.(x

$3)

2!

y2#

92.

x2!

(y!

5)2

#4

3.(x

$1)

2!

(y!

3)2

#9

(3,0

),r%

3(0

,!5)

,r%

2(1

,!3)

,r%

3

4.(x

$2)

2!

(y!

4)2

#16

5.x2

!y2

$10

x!

8y!

16 #

06.

x2!

y2$

4x!

6y#

12

(2,!

4),r

%4

(5,!

4),r

%5

(2,!

3),r

%5

x

y

Ox

y

Ox

y

O

x

y

Ox

y

O

x

y

O

x

y

O

x2 #

2x #

y2 #

4y %

11

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Circ

les

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-3

8-3

Exam

ple

Exam

ple

Exer

cises

Exer

cises

Page 57: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A9 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-3)

Skil

ls P

ract

ice

Circ

les

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-3

8-3

©G

lenc

oe/M

cGra

w-Hi

ll46

9G

lenc

oe A

lgeb

ra 2

Lesson 8-3

Wri

te a

n e

quat

ion

for

th

e ci

rcle

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

1.ce

nter

(0,

5),r

adiu

s 1

unit

2.ce

nter

(5,

12),

radi

us 8

uni

tsx2

#(y

!5)

2%

1(x

!5)

2#

(y!

12)2

%64

3.ce

nter

(4,

0),r

adiu

s 2

unit

s4.

cent

er (

2,2)

,rad

ius

3 un

its

(x!

4)2

#y2

%4

(x!

2)2

#(y

!2)

2%

9

5.ce

nter

(4,

$4)

,rad

ius

4 un

its

6.ce

nter

($6,

4),r

adiu

s 5

unit

s(x

!4)

2#

(y#

4)2

%16

(x#

6)2

#(y

!4)

2%

257.

endp

oint

s of

a d

iam

eter

at

($12

,0)

and

(12,

0)x2

#y2

%14

48.

endp

oint

s of

a d

iam

eter

at

($4,

0) a

nd ($

4,$

6)(x

#4)

2#

(y#

3)2

%9

9.ce

nter

at

(7,$

3),p

asse

s th

roug

h th

e or

igin

(x!

7)2

#(y

#3)

2%

5810

.cen

ter

at ($

4,4)

,pas

ses

thro

ugh

($4,

1)(x

#4)

2#

(y!

4)2

%9

11.c

ente

r at

($6,

$5)

,tan

gent

to

y-ax

is(x

#6)

2#

(y#

5)2

%36

12.c

ente

r at

(5,

1),t

ange

nt t

o x-

axis

(x!

5)2

#(y

!1)

2%

1

Fin

d t

he

cen

ter

and

rad

ius

of t

he

circ

le w

ith

th

e gi

ven

equ

atio

n.T

hen

gra

ph

th

eci

rcle

.

13.x

2!

y2#

914

.(x

$1)

2!

(y$

2)2

#4

15.(

x!

1)2

!y2

#16

(0,0

),3

units

(1,2

),2

units

(!1,

0),4

uni

ts

16.x

2!

(y!

3)2

#81

17.(

x$

5)2

!(y

!8)

2#

4918

.x2

!y2

$4y

$32

#0

(0,!

3),9

uni

ts(5

,!8)

,7 u

nits

(0,2

),6

units

x

y

O4

8

8 4 –4 –8

–4–8

x

y

O4

812

–4 –8 –12

x

y

O6

12

12 6 –6 –12

–6–1

2

x

y

Ox

y

Ox

y

O

©G

lenc

oe/M

cGra

w-Hi

ll47

0G

lenc

oe A

lgeb

ra 2

Wri

te a

n e

quat

ion

for

th

e ci

rcle

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

1.ce

nter

($4,

2),r

adiu

s 8

unit

s2.

cent

er (

0,0)

,rad

ius

4 un

its

(x#

4)2

#(y

!2)

2%

64x2

#y2

%16

3.ce

nter

!$,$

$3# ",

radi

us 5

$2#

unit

s4.

cent

er (

2.5,

4.2)

,rad

ius

0.9

unit

"x#

#2#

( y#

$3!)

2%

50(x

!2.

5)2

#(y

!4.

2)2

%0.

815.

endp

oint

s of

a d

iam

eter

at

($2,

$9)

and

(0,

$5)

(x#

1)2

#(y

#7)

2%

56.

cent

er a

t ($

9,$

12),

pass

es t

hrou

gh ($

4,$

5)(x

#9)

2#

(y#

12)2

%74

7.ce

nter

at

($6,

5),t

ange

nt t

o x-

axis

(x#

6)2

#(y

!5)

2%

25

Fin

d t

he

cen

ter

and

rad

ius

of t

he

circ

le w

ith

th

e gi

ven

equ

atio

n.T

hen

gra

ph

th

eci

rcle

.

8.(x

!3)

2!

y2#

169.

3x2

!3y

2#

1210

.x2

!y2

!2x

!6y

#26

(!3,

0),4

uni

ts(0

,0),

2 un

its(!

1,!

3),6

uni

ts

11.(

x $

1)2

!y2

!4y

#12

12.x

2$

6x!

y2#

013

.x2

!y2

!2x

!6y

#$

1(1

,!2)

,4 u

nits

(3,0

),3

units

(!1,

!3)

,3 u

nits

WEA

THER

For

Exe

rcis

es 1

4 an

d 1

5,u

se t

he

foll

owin

g in

form

atio

n.

On

aver

age,

the

circ

ular

eye

of a

hur

rica

ne is

abo

ut 1

5 m

iles

in d

iam

eter

.Gal

e w

inds

can

affe

ct a

n ar

ea u

p to

300

mile

s fr

om t

he s

torm

’s ce

nter

.In

1992

,Hur

rica

ne A

ndre

w d

evas

tate

dso

uthe

rn F

lori

da.A

sat

ellit

e ph

oto

of A

ndre

w’s

land

fall

show

ed t

he c

ente

r of

its

eye

on o

neco

ordi

nate

sys

tem

cou

ld b

e ap

prox

imat

ed b

y th

e po

int

(80,

26).

14.W

rite

an

equa

tion

to

repr

esen

t a

poss

ible

bou

ndar

y of

And

rew

’s e

ye.

(x!

80)2

#(y

!26

)2%

56.2

515

.Wri

te a

n eq

uati

on t

o re

pres

ent

a po

ssib

le b

ound

ary

of t

he a

rea

affe

cted

by

gale

win

ds.

(x!

80)2

#(y

!26

)2%

90,0

00

x

y

O

x

y

O

x

y

O

x

y

O4

8

4 –4 –8

–4–8

x

y

Ox

y

O

1 " 4

1 " 4

Pra

ctic

e (A

vera

ge)

Circ

les

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-3

8-3

Page 58: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A10 Glencoe Algebra 2

Answers (Lesson 8-3)

Rea

din

g t

o L

earn

Math

emati

csC

ircle

s

NAM

E__

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____

____

____

____

____

____

____

____

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____

DATE

____

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PERI

OD

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_

8-3

8-3

©G

lenc

oe/M

cGra

w-Hi

ll47

1G

lenc

oe A

lgeb

ra 2

Lesson 8-3

Pre-

Act

ivit

yW

hy

are

circ

les

imp

orta

nt

in a

ir t

raff

ic c

ontr

ol?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

3 at

the

top

of

page

426

in y

our

text

book

.

A la

rge

hom

e im

prov

emen

t ch

ain

is p

lann

ing

to e

nter

a n

ew m

etro

polit

anar

ea a

nd n

eeds

to

sele

ct lo

cati

ons

for

its

stor

es.M

arke

t re

sear

ch h

as s

how

nth

at p

oten

tial

cus

tom

ers

are

will

ing

to t

rave

l up

to 1

2 m

iles

to s

hop

at o

neof

the

ir s

tore

s.H

ow c

an c

ircl

es h

elp

the

man

ager

s de

cide

whe

re t

o pl

ace

thei

r st

ore?

Sam

ple

answ

er:A

sto

re w

ill d

raw

cus

tom

ers

who

live

insi

de a

circ

le w

ith c

ente

r at t

he s

tore

and

a ra

dius

of

12 m

iles.

The

man

agem

ent s

houl

d se

lect

loca

tions

for w

hich

as m

any

peop

le a

s po

ssib

le li

ve w

ithin

a c

ircle

of r

adiu

s 12

mile

s ar

ound

one

of t

he s

tore

s.

Rea

din

g t

he

Less

on

1.a.

Wri

te t

he e

quat

ion

of t

he c

ircl

e w

ith

cent

er (h

,k)

and

radi

us r

.(x

!h)

2#

(y!

k)2

%r2

b.W

rite

the

equ

atio

n of

the

cir

cle

wit

h ce

nter

(4,

$3)

and

rad

ius

5.(x

!4)

2#

(y#

3)2

%25

c.T

he c

ircl

e w

ith

equa

tion

(x !

8)2

!y2

#12

1 ha

s ce

nter

an

d ra

dius

.

d.T

he c

ircl

e w

ith

equa

tion

(x$

10)2

!(y

!10

)2#

1 ha

s ce

nter

an

d

radi

us

.

2.a.

In o

rder

to

find

cent

er a

nd r

adiu

s of

the

cir

cle

wit

h eq

uati

on x

2!

y2!

4x$

6y$

3 #

0,

it is

nec

essa

ry t

o .F

ill in

the

mis

sing

par

ts o

f th

ispr

oces

s.

x2!

y2!

4x$

6y$

3 #

0

x2!

y2!

4x$

6y#

x2!

4x!

!y2

$6y

!#

!!

(x!

)2!

(y$

)2#

b.T

his

circ

le h

as r

adiu

s 4

and

cent

er a

t .

Hel

pin

g Y

ou

Rem

emb

er

3.H

ow c

an t

he d

ista

nce

form

ula

help

you

to

rem

embe

r th

e eq

uati

on o

f a

circ

le?

Sam

ple

answ

er:W

rite

the

dist

ance

form

ula.

Repl

ace

(x1,

y 1) w

ith (h

,k)

and

(x2,

y 2) w

ith (x

,y).

Repl

ace

dw

ith r.

Squa

re b

oth

side

s.No

w y

ouha

ve th

e eq

uatio

n of

a c

ircle

.

(!2,

3)16

32

94

39

43

com

plet

e th

e sq

uare

1(1

0,!

10)

11(!

8,0)

©G

lenc

oe/M

cGra

w-Hi

ll47

2G

lenc

oe A

lgeb

ra 2

Tang

ents

to C

ircle

sA

line

tha

t in

ters

ects

a c

ircl

e in

exa

ctly

one

poi

nt is

a

tan

gen

tto

the

cir

cle.

In t

he d

iagr

am,l

ine

!is

ta

ngen

t to

the

cir

cle

wit

h eq

uati

on x

2!

y2#

25 a

t th

e po

int

who

se c

oord

inat

es a

re (

3,4)

.

A li

ne is

tan

gent

to

a ci

rcle

at

a po

int

Pon

the

cir

cle

if a

nd o

nly

if t

he li

ne is

per

pend

icul

ar t

o th

e ra

dius

fr

om t

he c

ente

r of

the

cir

cle

to p

oint

P.T

his

fact

en

able

s yo

u to

fin

d an

equ

atio

n of

the

tan

gent

to

a ci

rcle

at

a po

int

Pif

you

kno

w a

n eq

uati

on f

or t

he

circ

le a

nd t

he c

oord

inat

es o

f P.

Use

th

e d

iagr

am a

bove

to

solv

e ea

ch p

robl

em.

1.W

hat

is t

he s

lope

of

the

radi

us t

o th

e po

int

wit

h co

ordi

nate

s (3

,4)?

Wha

t is

the

slop

e of

the

tan

gent

to

that

poi

nt?

"4 3" ,!

"3 4"

2.F

ind

an e

quat

ion

of t

he li

ne !

that

is t

ange

nt t

o th

e ci

rcle

at

(3,4

).

y%

!"3 4" x

#"2 45 "

3.If

kis

a r

eal n

umbe

r be

twee

n $

5 an

d 5,

how

man

y po

ints

on

the

circ

le h

ave

x-co

ordi

nate

k?

Stat

e th

e co

ordi

nate

s of

the

se p

oint

s in

ter

ms

of k

.

two,

( k,(

$25

!!

k2 !)

4.D

escr

ibe

how

you

can

fin

d eq

uati

ons

for

the

tang

ents

to

the

poin

ts y

ou n

amed

for

Exe

rcis

e 3.

Use

the

coor

dina

tes

of (0

,0) a

nd o

f one

of t

he g

iven

poi

nts.

Find

the

slop

e of

the

radi

us to

that

poi

nt.U

se th

e sl

ope

of th

e ra

dius

to fi

nd w

hat

the

slop

e of

the

tang

ent m

ust b

e.Us

e th

e sl

ope

of th

e ta

ngen

t and

the

coor

dina

tes

of th

e po

int o

n th

e ci

rcle

to fi

nd a

n eq

uatio

n fo

r the

tang

ent.

5.F

ind

an e

quat

ion

for

the

tang

ent

at ($

3,4)

.

y%

"3 4" x#

"2 x5 "

5

–5

–5

5

(3, 4

)

y

xO

!x2

# y

2 % 2

5

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-3

8-3

Page 59: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A11 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-4)

Stu

dy

Gu

ide

and I

nte

rven

tion

Ellip

ses

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-4

8-4

©G

lenc

oe/M

cGra

w-Hi

ll47

3G

lenc

oe A

lgeb

ra 2

Lesson 8-4

Equ

atio

ns

of

Ellip

ses

An

elli

pse

is t

he s

et o

f al

l poi

nts

in a

pla

ne s

uch

that

the

sum

of t

he d

ista

nces

fro

m t

wo

give

n po

ints

in t

he p

lane

,cal

led

the

foci

,is

cons

tant

.An

ellip

seha

s tw

o ax

es o

f sy

mm

etry

whi

ch c

onta

in t

he m

ajor

and

min

or a

xes.

In t

he t

able

,the

leng

ths

a,b,

and

car

e re

late

d by

the

for

mul

a c2

#a2

$ b

2 .

Stan

dard

For

m o

f Equ

atio

n!

#1

!#

1

Cent

er(h

, k)

(h, k

)

Dire

ctio

n of

Maj

or A

xis

Horiz

onta

lVe

rtica

l

Foci

(h!

c, k

), (h

$c,

k)

(h, k

$c)

, (h,

k!

c)

Leng

th o

f Maj

or A

xis

2aun

its2a

units

Leng

th o

f Min

or A

xis

2bun

its2b

units

Wri

te a

n e

quat

ion

for

th

e el

lip

se s

how

n.

The

leng

th o

f th

e m

ajor

axi

s is

the

dis

tanc

e be

twee

n ($

2,$

2)

and

($2,

8).T

his

dist

ance

is 1

0 un

its.

2a#

10,s

o a

#5

The

foc

i are

loca

ted

at ($

2,6)

and

($2,

0),s

o c

#3.

b2#

a2$

c2#

25 $

9#

16T

he c

ente

r of

the

elli

pse

is a

t ($

2,3)

,so

h#

$2,

k#

3,a2

#25

,and

b2

#16

.The

maj

or a

xis

is v

erti

cal.

An

equa

tion

of

the

ellip

se is

!

#1.

Wri

te a

n e

quat

ion

for

th

e el

lip

se t

hat

sat

isfi

es e

ach

set

of

con

dit

ion

s.

1.en

dpoi

nts

of m

ajor

axi

s at

($7,

2) a

nd (5

,2),

endp

oint

s of

min

or a

xis

at ($

1,0)

and

($1,

4)

#%

1

2.m

ajor

axi

s 8

unit

s lo

ng a

nd p

aral

lel t

o th

e x-

axis

,min

or a

xis

2 un

its

long

,cen

ter

at ($

2,$

5)

#(y

#5)

2%

1

3.en

dpoi

nts

of m

ajor

axi

s at

($8,

4) a

nd (

4,4)

,foc

i at

($3,

4) a

nd ($

1,4)

#%

1

4.en

dpoi

nts

of m

ajor

axi

s at

(3,2

) and

(3,$

14),

endp

oint

s of

min

or a

xis

at ($

1,$

6) a

nd (7

,$6)

#%

1

5.m

inor

axi

s 6

unit

s lo

ng a

nd p

aral

lel t

o th

e x-

axis

,maj

or a

xis

12 u

nits

long

,cen

ter

at (6

,1)

#%

1(x

!6)

2"

9(y

!1)

2"

36

(x!

3)2

"16

(y#

6)2

"64

(y!

4)2

"35

(x#

2)2

"36

(x#

2)2

"16

(y!

2)2

"4

(x#

1)2

"36

(x!

2)2

"16

(y$

3)2

"25

x

F 1 F 2O

y

(x$

h)2

"b2

(y$

k)2

"a2

(y$

k)2

"b2

(x$

h)2

"a2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll47

4G

lenc

oe A

lgeb

ra 2

Gra

ph

Elli

pse

sTo

gra

ph a

n el

lipse

,if

nece

ssar

y,w

rite

the

giv

en e

quat

ion

in t

hest

anda

rd f

orm

of

an e

quat

ion

for

an e

llips

e.

!#

1 (f

or e

llips

e w

ith

maj

or a

xis

hori

zont

al)

or

!#

1 (f

or e

llips

e w

ith

maj

or a

xis

vert

ical

)

Use

the

cen

ter

(h,k

) an

d th

e en

dpoi

nts

of t

he a

xes

to p

lot

four

poi

nts

of t

he e

llips

e.To

mak

ea

mor

e ac

cura

te g

raph

,use

a c

alcu

lato

r to

fin

d so

me

appr

oxim

ate

valu

es f

or x

and

yth

atsa

tisf

y th

e eq

uati

on.

Gra

ph

th

e el

lip

se 4

x2#

6y2

#8x

!36

y%

!34

.

4x2

!6y

2!

8x$

36y

#$

344x

2!

8x!

6y2

$36

y#

$34

4(x2

!2x

!■

) !6(

y2$

6y!

■) #

$34

!■

4(x2

!2x

!1)

!6(

y2$

6y!

9) #

$34

!58

4(x

!1)

2!

6(y

$3)

2#

24

!#

1

The

cen

ter

of t

he e

llips

e is

($1,

3).S

ince

a2

#6,

a#

$6#.

Sinc

e b2

#4,

b#

2.T

he le

ngth

of

the

maj

or a

xis

is 2

$6#,

and

the

leng

th o

f th

e m

inor

axi

s is

4.S

ince

the

x-t

erm

has

the

grea

ter

deno

min

ator

,the

maj

or a

xis

is h

oriz

onta

l.P

lot

the

endp

oint

s of

the

axe

s.T

hen

grap

h th

e el

lipse

.

Fin

d t

he

coor

din

ates

of

the

cen

ter

and

th

e le

ngt

hs

of t

he

maj

or a

nd

min

or a

xes

for

the

elli

pse

wit

h t

he

give

n e

quat

ion

.Th

en g

rap

h t

he

elli

pse

.

1.!

#1

(0,0

),4$

3!,6

2.!

#1

(0,0

),10

,4

3.x2

!4y

2!

24y

#$

32(0

,!3)

,4,2

4.9x

2!

6y2

$36

x!

12y

#12

(2,!

1),6

,2$

6!

x

y

Ox

y

O

x

y

Ox

y

O

y2" 4

x2" 25

x2" 9

y2" 12

(y$

3)2

"4

(x!

1)2

"6

xO

y

4x2 #

6y2

# 8

x ! 3

6y %

!34

(x$

h)2

"b2

(y$

k)2

"a2

(y$

k)2

"b2

(x$

h)2

"a2

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Ellip

ses

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-4

8-4

Exam

ple

Exam

ple

Exer

cises

Exer

cises

Page 60: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

©G

lencoe/McG

raw-HillA12

Glencoe Algebra 2

Answers

(Lesson 8-4)

Skills PracticeEllipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

© Glencoe/McGraw-Hill 475 Glencoe Algebra 2

Less

on

8-4

Write an equation for each ellipse.

1. 2. 3.

# % 1 # % 1 # % 1

Write an equation for the ellipse that satisfies each set of conditions.

4. endpoints of major axis 5. endpoints of major axis 6. endpoints of major axis at (0, 6) and (0, $6), at (2, 6) and (8, 6), at (7, 3) and (7, 9),endpoints of minor axis endpoints of minor axis endpoints of minor axis at ($3, 0) and (3, 0) at (5, 4) and (5, 8) at (5, 6) and (9, 6)

# % 1 # % 1 # % 1

7. major axis 12 units long 8. endpoints of major axis 9. endpoints of major axis atand parallel to x-axis, at ($6, 0) and (6, 0), foci (0, 12) and (0, $12), foci atminor axis 4 units long, at ($$32#, 0) and ($32#, 0) (0, $23# ) and (0, $$23# )center at (0, 0)

# % 1 # % 1 # % 1

Find the coordinates of the center and foci and the lengths of the major andminor axes for the ellipse with the given equation. Then graph the ellipse.

10. ! # 1 11. ! # 1 12. ! # 1

(0, 0); (0, ($19!); (0, 0); ((6$2!, 0); (0, 0), (0, (2$6!);20; 18 18; 6 14; 10

x

y

O 4 8

8

4

–4

–8

–4–8x

y

O 4 8

8

4

–4

–8

–4–8x

y

O 4 8

8

4

–4

–8

–4–8

x2"25

y2"49

y2"9

x2"81

x2"81

y2"100

x2"121

y2"144

y2"4

x2"36

y2"4

x2"36

(x ! 7)2"4

(y ! 6)2"9

(y ! 6)2"4

(x ! 5)2"9

x2"9

y2"36

(y ! 2)2"9

x2"16

x2"16

y2"25

y2"4

x2"9

xO

y(0, 5)

(0, –1)

(–4, 2) (4, 2)

xO

y

(0, 3)

(0, –3)

(0, –5)

(0, 5)

xO

y

(0, 2)

(0, –2)

(–3, 0)(3, 0)

© Glencoe/McGraw-Hill 476 Glencoe Algebra 2

Write an equation for each ellipse.

1. 2. 3.

# % 1 # % 1 # % 1

Write an equation for the ellipse that satisfies each set of conditions.

4. endpoints of major axis 5. endpoints of major axis 6. major axis 20 units long at ($9, 0) and (9, 0), at (4, 2) and (4, $8), and parallel to x-axis,endpoints of minor axis endpoints of minor axis minor axis 10 units long,at (0, 3) and (0, $3) at (1, $3) and (7, $3) center at (2, 1)

# % 1 # % 1 # % 1

7. major axis 10 units long, 8. major axis 16 units long, 9. endpoints of minor axis minor axis 6 units long center at (0, 0), foci at at (0, 2) and (0, $2), foci and parallel to x-axis, (0, 2$15# ) and (0, $2$15# ) at ($4, 0) and (4, 0)center at (2, $4)

# % 1 # % 1 # % 1

Find the coordinates of the center and foci and the lengths of the major andminor axes for the ellipse with the given equation. Then graph the ellipse.

10. ! # 1 11. ! # 1 12. ! # 1

(0, 0); (0, ($7!); 8; 6 (3, 1); (3, 1 ( $35! ); (!4, !3);12; 2 (!4 ( 2$6!, !3); 14; 10

13. SPORTS An ice skater traces two congruent ellipses to form a figure eight. Assume that thecenter of the first loop is at the origin, with the second loop to its right. Write an equationto model the first loop if its major axis (along the x-axis) is 12 feet long and its minoraxis is 6 feet long. Write another equation to model the second loop.

# % 1; # % 1y2"9

(x ! 12)2""36

y2"9

x2"36

4

4

–4

–8

–12

–4–8 x

y

O

x

y

O 4 8

8

4

–4

–8

–4–8x

y

O

( y ! 3)2"25

(x ! 4)2"49

(x $ 3)2"1

( y $ 1)2"36

x2"9

y2"16

y2"4

x2"20

x2"4

y2"64

(x ! 2)2"9

(y # 4)2"25

(y ! 1)2"25

(x ! 2)2"100

(x ! 4)2"9

(y # 3)2"25

y2"9

x2"81

(y ! 3)2"9

(x # 1)2"25

x2"4

(y ! 2)2"9

y2"9

x2"121

xO

y

(–5, 3)

(–6, 3)

(3, 3)

(4, 3)

xO

y

(0, 2 ! $%5)

(0, 2 # $%5)

(0, –1)

(0, 5)

xO

y(0, 3)

(0, –3)

(–11, 0) (11, 0)6 12

2

–2

–6–12

Practice (Average)

Ellipses

NAME ______________________________________________ DATE ____________ PERIOD _____

8-48-4

Page 61: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A13 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-4)

Rea

din

g t

o L

earn

Math

emati

csEl

lipse

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-4

8-4

©G

lenc

oe/M

cGra

w-Hi

ll47

7G

lenc

oe A

lgeb

ra 2

Lesson 8-4

Pre-

Act

ivit

yW

hy

are

elli

pse

s im

por

tan

t in

th

e st

ud

y of

th

e so

lar

syst

em?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

4 at

the

top

of

page

433

in y

our

text

book

.

Is t

he E

arth

alw

ays

the

sam

e di

stan

ce f

rom

the

Sun

? E

xpla

in y

our

answ

erus

ing

the

wor

ds c

ircl

ean

d el

lips

e.No

;if t

he E

arth

’s o

rbit

wer

e a

circ

le,i

t wou

ld a

lway

s be

the

sam

e di

stan

ce fr

om th

e Su

nbe

caus

e ev

ery

poin

t on

a ci

rcle

is th

e sa

me

dist

ance

from

the

cent

er.H

owev

er,t

he E

arth

’s o

rbit

is a

n el

lipse

,and

the

poin

tson

an

ellip

se a

re n

ot a

ll th

e sa

me

dist

ance

from

the

cent

er.

Rea

din

g t

he

Less

on

1.A

n el

lipse

is t

he s

et o

f al

l poi

nts

in a

pla

ne s

uch

that

the

of

the

dist

ance

s fr

om t

wo

fixe

d po

ints

is

.The

tw

o fi

xed

poin

ts a

re c

alle

d th

e

of t

he e

llips

e.

2.C

onsi

der

the

ellip

se w

ith

equa

tion

!

#1.

a.Fo

r th

is e

quat

ion,

a#

and

b#

.

b.W

rite

an

equa

tion

tha

t re

late

s th

e va

lues

of a

,b,a

nd c

.c2

%a2

!b2

c.F

ind

the

valu

e of

cfo

r th

is e

llips

e.$

5!

3.C

onsi

der

the

ellip

ses

wit

h eq

uati

ons

!#

1 an

d !

#1.

Com

plet

e th

e

follo

win

g ta

ble

to d

escr

ibe

char

acte

rist

ics

of t

heir

gra

phs.

Stan

dard

For

m o

f Equ

atio

n!

#1

!#

1

Dire

ctio

n of

Maj

or A

xis

vert

ical

horiz

onta

l

Dire

ctio

n of

Min

or A

xis

horiz

onta

lve

rtic

al

Foci

(0,3

),(0

,!3)

( $5!,

0) ,( !

$5!,

0)Le

ngth

of M

ajor

Axi

s10

uni

ts6

units

Leng

th o

f Min

or A

xis

8 un

its4

units

Hel

pin

g Y

ou

Rem

emb

er4.

Som

e st

uden

ts h

ave

trou

ble

rem

embe

ring

the

tw

o st

anda

rd f

orm

s fo

r th

e eq

uati

on o

f an

ellip

se.H

ow c

an y

ou r

emem

ber

whi

ch t

erm

com

es f

irst

and

whe

re t

o pl

ace

a an

d b

inth

ese

equa

tion

s?Th

e x-

axis

is h

oriz

onta

l.If

the

maj

or a

xis

is h

oriz

onta

l,th

e fir

st te

rm is

.T

he y

-axi

s is

ver

tical

.If t

he m

ajor

axi

s is

ver

tical

,the

first

term

is

.ais

alw

ays

the

larg

er o

f the

num

bers

aan

d b.

y2" a2x2" a2

y2" 4

x2" 9

x2" 16

y2" 25

y2" 4

x2" 9

x2" 16

y2" 25

23

y2" 4

x2" 9

foci

cons

tant

sum

©G

lenc

oe/M

cGra

w-Hi

ll47

8G

lenc

oe A

lgeb

ra 2

Ecce

ntric

ity

In a

n el

lipse

,the

rat

io " dc "

is c

alle

d th

e ec

cen

tric

ity

and

is d

enot

ed b

y th

e

lett

er e

.Ecc

entr

icit

y m

easu

res

the

elon

gati

on o

f an

ellip

se.T

he c

lose

r e

is t

o 0,

the

mor

e an

elli

pse

look

s lik

e a

circ

le.T

he c

lose

r e

is t

o 1,

the

mor

e el

onga

ted

it is

.Rec

all t

hat

the

equa

tion

of

an e

llips

e is

" ax2 2"!

" by2 2"#

1 or

" bx2 2"!

" ay2 2"#

1

whe

re a

is t

he le

ngth

of

the

maj

or a

xis,

and

that

c#

$a2

$b

#2 #.

Fin

d t

he

ecce

ntr

icit

y of

eac

h e

llip

se r

oun

ded

to

the

nea

rest

hu

nd

red

th.

1."x 92 "

!" 3y 62 "

#1

2." 8x 12 "

!"y 92 "

#1

3."x 42 "

!"y 92 "

#1

0.87

0.94

0.75

4." 1x 62 "

!"y 92 "

#1

5." 3x 62 "

!" 1y 62 "

#1

6."x 42 "

!" 3y 62 "

#1

0.66

0.75

0.94

7.Is

a c

ircl

e an

elli

pse?

Exp

lain

you

r re

ason

ing.

Yes;

it is

an

ellip

se w

ith e

ccen

trici

ty 0

.

8.T

he c

ente

r of

the

sun

is o

ne f

ocus

of

Ear

th's

orb

it a

roun

d th

e su

n.T

hele

ngth

of

the

maj

or a

xis

is 1

86,0

00,0

00 m

iles,

and

the

foci

are

3,2

00,0

00m

iles

apar

t.F

ind

the

ecce

ntri

city

of

Ear

th's

orb

it.

appr

oxim

atel

y 0.

17

9.A

n ar

tifi

cial

sat

ellit

e or

biti

ng t

he e

arth

tra

vels

at

an a

ltit

ude

that

var

ies

betw

een

132

mile

s an

d 58

3 m

iles

abov

e th

e su

rfac

e of

the

ear

th.I

f th

ece

nter

of

the

eart

h is

one

foc

us o

f it

s el

lipti

cal o

rbit

and

the

rad

ius

of t

heea

rth

is 3

950

mile

s,w

hat

is t

he e

ccen

tric

ity

of t

he o

rbit

?

appr

oxim

atel

y 0.

052

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-4

8-4

Page 62: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A14 Glencoe Algebra 2

Answers (Lesson 8-5)

Stu

dy

Gu

ide

and I

nte

rven

tion

Hyp

erbo

las

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

©G

lenc

oe/M

cGra

w-Hi

ll47

9G

lenc

oe A

lgeb

ra 2

Lesson 8-5

Equ

atio

ns

of

Hyp

erb

ola

sA

hyp

erbo

lais

the

set

of

all p

oint

s in

a p

lane

suc

h th

atth

e ab

solu

te v

alue

of

the

diff

eren

ceof

the

dis

tanc

es f

rom

any

poi

nt o

n th

e hy

perb

ola

to a

nytw

o gi

ven

poin

ts in

the

pla

ne,c

alle

d th

e fo

ci,i

s co

nsta

nt.

In t

he t

able

,the

leng

ths

a,b,

and

car

e re

late

d by

the

for

mul

a c2

#a2

!b2

.

Stan

dard

For

m o

f Equ

atio

n$

#1

$#

1

Equa

tions

of t

he A

sym

ptot

esy

$k

#(

(x$

h)y

$k

#(

(x$

h)

Tran

sver

se A

xis

Horiz

onta

lVe

rtica

l

Foci

(h$

c, k

), (h

!c,

k)

(h, k

$c)

, (h,

k!

c)

Vert

ices

(h$

a, k

), (h

!a,

k)

(h, k

$a)

, (h,

k!

a)

Wri

te a

n e

quat

ion

for

th

e h

yper

bola

wit

h v

erti

ces

(!2,

1) a

nd

(6,

1)an

d f

oci

(!4,

1) a

nd

(8,

1).

Use

a s

ketc

h to

ori

ent

the

hype

rbol

a co

rrec

tly.

The

cen

ter

of

the

hype

rbol

a is

the

mid

poin

t of

the

seg

men

t jo

inin

g th

e tw

o

vert

ices

.The

cen

ter

is (

,1),

or (

2,1)

.The

val

ue o

f ais

the

dist

ance

fro

m t

he c

ente

r to

a v

erte

x,so

a#

4.T

he v

alue

of c

is

the

dist

ance

fro

m t

he c

ente

r to

a f

ocus

,so

c#

6.

c2#

a2!

b2

62#

42!

b2

b2#

36 $

16 #

20

Use

h,k

,a2 ,

and

b2to

wri

te a

n eq

uati

on o

f th

e hy

perb

ola.

$#

1

Wri

te a

n e

quat

ion

for

th

e h

yper

bola

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

1.ve

rtic

es ($

7,0)

and

(7,

0),c

onju

gate

axi

s of

leng

th 1

0!

%1

2.ve

rtic

es (

$2,

$3)

and

(4,

$3)

,foc

i ($

5,$

3) a

nd (

7,$

3)!

%1

3.ve

rtic

es (

4,3)

and

(4,

$5)

,con

juga

te a

xis

of le

ngth

4!

%1

4.ve

rtic

es (

$8,

0) a

nd (

8,0)

,equ

atio

n of

asy

mpt

otes

y#

(x

!%

1

5.ve

rtic

es (

$4,

6) a

nd ($

4,$

2),f

oci (

$4,

10)

and

($4,

$6)

!%

1(x

#4)

2"

48(y

!2)

2"

16

9y2

" 16x2 " 64

1 " 6

(x!

4)2

"4

(y#

1)2

"16

(y#

3)2

"27

(x!

1)2

"9

y2 " 25x2 " 49

(y$

1)2

"20

(x$

2)2

"16

$2

!6

"2

x

y

O

a " bb " a

(x$

h)2

"b2

(y$

k)2

"a2

(y$

k)2

"b2

(x$

h)2

"a2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll48

0G

lenc

oe A

lgeb

ra 2

Gra

ph

Hyp

erb

ola

sTo

gra

ph a

hyp

erbo

la,w

rite

the

giv

en e

quat

ion

in t

he s

tand

ard

form

of

an e

quat

ion

for

a hy

perb

ola

$#

1 if

the

bra

nche

s of

the

hyp

erbo

la o

pen

left

and

rig

ht,o

r

$#

1 if

the

bra

nche

s of

the

hyp

erbo

la o

pen

up a

nd d

own

Gra

ph t

he p

oint

(h,k

),w

hich

is t

he c

ente

r of

the

hyp

erbo

la.D

raw

a r

ecta

ngle

wit

hdi

men

sion

s 2a

and

2ban

d ce

nter

(h,k

).If

the

hyp

erbo

la o

pens

left

and

rig

ht,t

he v

erti

ces

are

(h$

a,k)

and

(h

!a,

k).I

f th

e hy

perb

ola

open

s up

and

dow

n,th

e ve

rtic

es a

re (h

,k$

a)an

d (h

,k!

a).

Dra

w t

he

grap

h o

f 6y

2!

4x2

!36

y!

8x%

!26

.

Com

plet

e th

e sq

uare

s to

get

the

equ

atio

n in

sta

ndar

d fo

rm.

6y2

$4x

2$

36y

$8x

#$

266(

y2$

6y!

■) $

4(x2

!2x

!■

) #$

26 !

■6(

y2$

6y!

9) $

4(x2

!2x

!1)

#$

26 !

506(

y$

3)2

$4(

x!

1)2

#24

$#

1

The

cen

ter

of t

he h

yper

bola

is ($

1,3)

.A

ccor

ding

to

the

equa

tion

,a2

#4

and

b2#

6,so

a#

2 an

d b

#$

6#.T

he t

rans

vers

e ax

is is

ver

tica

l,so

the

ver

tice

s ar

e ($

1,5)

and

($1,

1).D

raw

a r

ecta

ngle

wit

hve

rtic

al d

imen

sion

4 a

nd h

oriz

onta

l dim

ensi

on 2

$6#

%4.

9.T

he d

iago

nals

of

this

rec

tang

lear

e th

e as

ympt

otes

.The

bra

nche

s of

the

hyp

erbo

la o

pen

up a

nd d

own.

Use

the

ver

tice

s an

dth

e as

ympt

otes

to

sket

ch t

he h

yper

bola

.

Fin

d t

he

coor

din

ates

of

the

vert

ices

an

d f

oci

and

th

e eq

uat

ion

s of

th

e as

ymp

tote

sfo

r th

e h

yper

bola

wit

h t

he

give

n e

quat

ion

.Th

en g

rap

h t

he

hyp

erbo

la.

1.$

#1

2.(y

$3)

2$

#1

3.$

#1

(2,0

),(!

2,0)

;(!

2,4)

,(!

2,2)

;(0

,4),

(0,!

4);

( 2$5!,

0) ,( !

2$5!,

0) ;( !

2,3

#$

10!) ,

(0,5

),(0

,!5)

;y

%(

2x( !

2,3

!$

10!) ;

y%

(x

y %

x #

3,

y %

!x

#21 " 3

1 " 3

2 " 31 " 3

xO

y

4 " 3

x2" 9

y2" 16

(x!

2)2

"9

y2" 16

x2" 4

(x!

1)2

"6

(y$

3)2

"4

xO

y

(x$

h)2

"b2

(y$

k)2

"a2

(y$

k)2

"" b2

(x$

h)2

"a2

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Hyp

erbo

las

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

Exam

ple

Exam

ple

Exer

cises

Exer

cises

xO

y

xO

y

Page 63: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A15 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-5)

Skil

ls P

ract

ice

Hyp

erbo

las

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

©G

lenc

oe/M

cGra

w-Hi

ll48

1G

lenc

oe A

lgeb

ra 2

Lesson 8-5

Wri

te a

n e

quat

ion

for

eac

h h

yper

bola

.

1.2.

3.

!%

1!

%1

!%

1

Wri

te a

n e

quat

ion

for

th

e h

yper

bola

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

4.ve

rtic

es ($

4,0)

and

(4,

0),c

onju

gate

axi

s of

leng

th 8

!%

1

5.ve

rtic

es (

0,6)

and

(0,

$6)

,con

juga

te a

xis

of le

ngth

14

!%

1

6.ve

rtic

es (

0,3)

and

(0,

$3)

,con

juga

te a

xis

of le

ngth

10

!%

1

7.ve

rtic

es (

$2,

0) a

nd (

2,0)

,con

juga

te a

xis

of le

ngth

4!

%1

8.ve

rtic

es (

$3,

0) a

nd (

3,0)

,foc

i ((

5,0)

!%

1

9.ve

rtic

es (

0,2)

and

(0,

$2)

,foc

i (0,

(3)

!%

1

10.v

erti

ces

(0,$

2) a

nd (

6,$

2),f

oci (

3 (

$13#

,$2)

!%

1

Fin

d t

he

coor

din

ates

of

the

vert

ices

an

d f

oci

and

th

e eq

uat

ion

s of

th

e as

ymp

tote

sfo

r th

e h

yper

bola

wit

h t

he

give

n e

quat

ion

.Th

en g

rap

h t

he

hyp

erbo

la.

11.

$#

112

.$

#1

13.

$#

1

((3,

0);( (

3$5!,

0) ;(0

,(7)

;( 0,(

$58!

) ;((

4,0)

;( ($

17!,0

) ;y

%(

2xy

%(

xy

%(

x

xO

y

48

8 4 –4 –8

–4–8

xO

y

48

8 4 –4 –8

–4–8

xO

y

1 " 47 " 3

y2" 1

x2" 16

x2" 9

y2" 49

y2" 36

x2" 9

(y#

2)2

"4

(x!

3)2

"9

x2" 5

y2 " 4

y2 " 16x2" 9

y2 " 4x2" 4

x2" 25

y2 " 9

x2" 49

y2 " 36

y2 " 16x2" 16

y2" 25

x2" 4

x2" 25

y2 " 36y2 " 16

x2" 25

x

y

O

( $!29

, 0)

( –$

!29, 0

)

( 2, 0

)(–

2, 0

)

48

8 4 –4 –8

–4–8

x

y

O

( 0, $

!61)

( 0, –

$!61

)

( 0, 6

)

(0, –

6)48

8 4 –4 –8

–4–8

x

y

O

( $!41

, 0)

( –$

!41, 0

)

( 5, 0

)

(–5,

0)

48

8 4 –4 –8

–4–8

©G

lenc

oe/M

cGra

w-Hi

ll48

2G

lenc

oe A

lgeb

ra 2

Wri

te a

n e

quat

ion

for

eac

h h

yper

bola

.

1.2.

3.

!%

1!

%1

!%

1

Wri

te a

n e

quat

ion

for

th

e h

yper

bola

th

at s

atis

fies

eac

h s

et o

f co

nd

itio

ns.

4.ve

rtic

es (

0,7)

and

(0,

$7)

,con

juga

te a

xis

of le

ngth

18

unit

s!

%1

5.ve

rtic

es (

1,$

1) a

nd (

1,$

9),c

onju

gate

axi

s of

leng

th 6

uni

ts!

%1

6.ve

rtic

es ($

5,0)

and

(5,

0),f

oci (

($

26#,0

)!

%1

7.ve

rtic

es (

1,1)

and

(1,

$3)

,foc

i (1,

$1

($

5#)!

%1

Fin

d t

he

coor

din

ates

of

the

vert

ices

an

d f

oci

and

th

e eq

uat

ion

s of

th

e as

ymp

tote

sfo

r th

e h

yper

bola

wit

h t

he

give

n e

quat

ion

.Th

en g

rap

h t

he

hyp

erbo

la.

8.$

#1

9.$

#1

10.

$#

1

(0,(

4);( 0

,(2$

5!) ;(1

,3),

(1,1

);(3

,0),

(3,!

4);

y%

(2x

( 1,2

($

5!) ;( 3,

!2

(2$

2!) ;y

!2

%(

(x!

1)y

#2

%(

(x!

3)

11.A

STR

ON

OM

YA

stro

nom

ers

use

spec

ial X

-ray

tel

esco

pes

to o

bser

ve t

he s

ourc

es o

fce

lest

ial X

ray

s.So

me

X-r

ay t

eles

cope

s ar

e fi

tted

wit

h a

met

al m

irro

r in

the

sha

pe o

f a

hype

rbol

a,w

hich

ref

lect

s th

e X

ray

s to

a f

ocus

.Sup

pose

the

ver

tice

s of

suc

h a

mir

ror

are

loca

ted

at ($

3,0)

and

(3,

0),a

nd o

ne f

ocus

is lo

cate

d at

(5,

0).W

rite

an

equa

tion

tha

tm

odel

s th

e hy

perb

ola

form

ed b

y th

e m

irro

r.!

%1

y2" 16

x2" 9

xO

y

xO

y

xO

y

48

8 4 –4 –8

–4–8

1 " 2

(x$

3)2

"4

(y!

2)2

"4

(x $

1)2

"4

(y$

2)2

"1

x2" 4

y2" 16

(x!

1)2

"1

(y#

1)2

"4y2" 1

x2" 25

(x!

1)2

"9

(y#

5)2

"16

x2" 81

y2 " 49

(y#

2)2

"16

(x !

1)2

"4

(x #

3)2

"25

(y !

2)2

"9

x2" 36

y2 " 9

x

y

O(–

1, –

2)

(1, –

2)

(3, –

2)x

y O

( –3,

2 #

$!34

)

( –3,

2 !

$!34

)

( –3,

–1)

(–3,

5)

4

8 4 –4

–4–8

x

y

O

( 0, 3

$%5)

( 0, –

3$%5)

( 0, 3

)

(0, –

3)48

8 4 –4 –8

–4–8

Pra

ctic

e (A

vera

ge)

Hyp

erbo

las

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

Page 64: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A16 Glencoe Algebra 2

Answers (Lesson 8-5)

Rea

din

g t

o L

earn

Math

emati

csH

yper

bola

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

©G

lenc

oe/M

cGra

w-Hi

ll48

3G

lenc

oe A

lgeb

ra 2

Lesson 8-5

Pre-

Act

ivit

yH

ow a

re h

yper

bola

s d

iffe

ren

t fr

om p

arab

olas

?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

5 at

the

top

of

page

441

in y

our

text

book

.

Loo

k at

the

ske

tch

of a

hyp

erbo

la in

the

intr

oduc

tion

to

this

less

on.L

ist

thre

e w

ays

in w

hich

hyp

erbo

las

are

diff

eren

t fr

om p

arab

olas

.Sa

mpl

e an

swer

:A h

yper

bola

has

two

bran

ches

,whi

le a

para

bola

is o

ne c

ontin

uous

cur

ve.A

hyp

erbo

la h

as tw

o fo

ci,

whi

le a

par

abol

a ha

s on

e fo

cus.

A hy

perb

ola

has

two

vert

ices

,w

hile

a p

arab

ola

has

one

vert

ex.

Rea

din

g t

he

Less

on

1.T

he g

raph

at

the

righ

t sh

ows

the

hype

rbol

a w

hose

equa

tion

in s

tand

ard

form

is

$#

1.

The

poi

nt (

0,0)

is t

he

of t

he

hype

rbol

a.

The

poi

nts

(4,0

) an

d ($

4,0)

are

the

of

the

hyp

erbo

la.

The

poi

nts

(5,0

) an

d ($

5,0)

are

the

of

the

hyp

erbo

la.

The

seg

men

t co

nnec

ting

(4,

0) a

nd ($

4,0)

is c

alle

d th

e ax

is.

The

seg

men

t co

nnec

ting

(0,

3) a

nd (

0,$

3) is

cal

led

the

axis

.

The

line

s y

#x

and

y#

$x

are

calle

d th

e .

2.St

udy

the

hype

rbol

a gr

aphe

d at

the

rig

ht.

The

cen

ter

is

.

The

val

ue o

f ais

.

The

val

ue o

f cis

.

To fi

nd b

2 ,so

lve

the

equa

tion

#

!.

The

equ

atio

n in

sta

ndar

d fo

rm f

or t

his

hype

rbol

a is

.

Hel

pin

g Y

ou

Rem

emb

er

3.W

hat

is a

n ea

sy w

ay t

o re

mem

ber

the

equa

tion

rel

atin

g th

e va

lues

of a

,b,a

nd c

for

ahy

perb

ola?

This

equ

atio

n lo

oks

just

like

the

Pyth

agor

ean

Theo

rem

,al

thou

gh th

e va

riabl

es re

pres

ent d

iffer

ent l

engt

hs in

a h

yper

bola

than

ina

right

tria

ngle

.

"x 42 "!

" 1y 22 "%

1

b2a2

c242

(0,0

)

x

y

O

asym

ptot

es3 " 4

3 " 4

conj

ugat

etra

nsve

rse

foci

vert

ices

cent

er

y2" 9

x2" 16

x

y

O( –

4, 0

)( 4

, 0)

( –5,

0)

( 5, 0

)

y % 3 4x

y % –

3 4x

©G

lenc

oe/M

cGra

w-Hi

ll48

4G

lenc

oe A

lgeb

ra 2

Rec

tang

ular

Hyp

erbo

las

A r

ecta

ngu

lar

hyp

erbo

lais

a h

yper

bola

wit

h pe

rpen

dicu

lar

asym

ptot

es.

For

exam

ple,

the

grap

h of

x2

$y2

#1

is a

rec

tang

ular

hyp

erbo

la.A

hyp

erbo

law

ith

asym

ptot

es t

hat

are

not

perp

endi

cula

r is

cal

led

a n

onre

ctan

gula

rh

yper

bola

.The

gra

phs

of e

quat

ions

of

the

form

xy

#c,

whe

re c

is a

con

stan

t,ar

e re

ctan

gula

r hy

perb

olas

.

Mak

e a

tabl

e of

val

ues

an

d p

lot

poi

nts

to

grap

h e

ach

rec

tan

gula

rh

yper

bola

bel

ow.B

e su

re t

o co

nsi

der

neg

ativ

e va

lues

for

th

eva

riab

les.

See

stud

ents

’tab

les.

1.xy

#$

42.

xy#

3

3.xy

#$

14.

xy#

8

5.M

ake

a co

njec

ture

abo

ut t

he a

sym

ptot

es o

f re

ctan

gula

r hy

perb

olas

.

The

coor

dina

te a

xes

are

the

asym

ptot

es.

x

y

Ox

y

O

x

y

Ox

y

O

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-5

8-5

Page 65: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A17 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-6)

Stu

dy

Gu

ide

and I

nte

rven

tion

Con

ic S

ectio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

©G

lenc

oe/M

cGra

w-Hi

ll48

5G

lenc

oe A

lgeb

ra 2

Lesson 8-6

Stan

dar

d F

orm

Any

con

ic s

ecti

on in

the

coo

rdin

ate

plan

e ca

n be

des

crib

ed b

y an

equa

tion

of

the

form

A

x2!

Bxy

!C

y2!

Dx

!E

y!

F#

0,w

here

A,B

,and

Car

e no

t al

l zer

o.O

ne w

ay t

o te

ll w

hat

kind

of

coni

c se

ctio

n an

equ

atio

n re

pres

ents

is t

o re

arra

nge

term

s an

dco

mpl

ete

the

squa

re,i

f ne

cess

ary,

to g

et o

ne o

f th

e st

anda

rd f

orm

s fr

om a

n ea

rlie

r le

sson

.T

his

met

hod

is e

spec

ially

use

ful i

f yo

u ar

e go

ing

to g

raph

the

equ

atio

n.

Wri

te t

he

equ

atio

n 3

x2!

4y2

!30

x!

8y

#59

%0

in s

tan

dar

d f

orm

.S

tate

wh

eth

er t

he

grap

h o

f th

e eq

uat

ion

is

a pa

rabo

la,c

ircl

e,el

lips

e,or

hyp

erbo

la.

3x2

$4y

2$

30x

$ 8

y!

59#

0O

rigin

al e

quat

ion

3x2

$30

x$

4y2

$8y

#$

59Is

olat

e te

rms.

3(x2

$10

x!

■) $

4(y2

!2y

!■

)#

$59

! ■

!■

Fact

or o

ut c

omm

on m

ultip

les.

3(x2

$10

x!

25)

$4(

y2!

2y!

1)#

$59

!3(

25)

! ($

4)(1

)Co

mpl

ete

the

squa

res.

3(x

$5)

2$

4(y

!1)

2#

12Si

mpl

ify.

$#

1Di

vide

each

sid

e by

12.

The

gra

ph o

f th

e eq

uati

on is

a h

yper

bola

wit

h it

s ce

nter

at

(5,$

1).T

he le

ngth

of

the

tran

sver

se a

xis

is 4

uni

ts a

nd t

he le

ngth

of

the

conj

ugat

e ax

is is

2$

3#un

its.

Wri

te e

ach

equ

atio

n i

n s

tan

dar

d f

orm

.Sta

te w

het

her

th

e gr

aph

of

the

equ

atio

n i

sa

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.

1.x2

!y2

$6x

!4y

!3

#0

2.x2

!2y

2!

6x$

20y

!53

#0

(x!

3)2

#(y

#2)

2%

10;c

ircle

#%

1;el

lipse

3.6x

2$

60x

$y

!16

1 #

04.

x2!

y2$

4x$

14y

!29

#0

y%

6(x

!5)

2#

11;p

arab

ola

(x!

2)2

#(y

!7)

2%

24;c

ircle

5.6x

2$

5y2

!24

x!

20y

$56

#0

6.3y

2!

x$

24y

!46

#0

!%

1;hy

perb

ola

x%

!3(

y!

4)2

#2;

para

bola

7.x2

$4y

2$

16x

!24

y$

36 #

08.

x2!

2y2

!8x

!4y

!2

#0

!%

1;hy

perb

ola

#%

1;el

lipse

9.4x

2!

48x

!y

!15

8 #

010

.3x2

!y2

$48

x$

4y!

184

#0

y%

!4(

x#

6)2

!14

;par

abol

a#

%1;

ellip

se

11.$

3x2

!2y

2$

18x

!20

y!

5 #

012

.x2

!y2

!8x

!2y

!8

#0

!%

1;hy

perb

ola

(x#

4)2

#(y

#1)

2%

9;ci

rcle

(x#

3)2

"6

(y#

5)2

"9

(y!

2)2

"12

(x!

8)2

"4

(y#

1)2

"8

(x#

4)2

"16

(y!

3)2

"16

(x!

8)2

"64

(y!

2)2

"12

(x#

2)2

"10

(y!

5)2

"3

(x#

3)2

"6

(y!

1)2

"3

(x$

5)2

"4

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll48

6G

lenc

oe A

lgeb

ra 2

Iden

tify

Co

nic

Sec

tio

ns

If y

ou a

re g

iven

an

equa

tion

of

the

form

Ax2

!B

xy!

Cy2

!D

x!

Ey

!F

#0,

wit

h B

#0,

you

can

dete

rmin

e th

e ty

pe o

f co

nic

sect

ion

just

by

cons

ider

ing

the

valu

es o

f Aan

d C

.Ref

erto

the

fol

low

ing

char

t.

Rela

tions

hip

of A

and

CTy

pe o

f Con

ic S

ectio

n

A#

0 or

C#

0, b

ut n

ot b

oth.

para

bola

A #

Ccir

cle

Aan

d C

have

the

sam

e sig

n, b

ut A

)C

.el

lipse

Aan

d C

have

opp

osite

sig

ns.

hype

rbol

a

Wit

hou

t w

riti

ng

the

equ

atio

n i

n s

tan

dar

d f

orm

,sta

te w

het

her

th

egr

aph

of

each

equ

atio

n i

s a

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Con

ic S

ectio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

Exam

ple

Exam

ple

a.3x

2!

3y2

#5x

#12

%0

A#

3 an

d C

#$

3 ha

ve o

ppos

ite

sign

s,so

the

grap

h of

the

equ

atio

n is

a h

yper

bola

.

b.y2

%7y

!2x

#13

A#

0,so

the

gra

ph o

f th

e eq

uati

on is

a pa

rabo

la.

Exer

cises

Exer

cises

Wit

hou

t w

riti

ng

the

equ

atio

n i

n s

tan

dar

d f

orm

,sta

te w

het

her

th

e gr

aph

of

each

equ

atio

n i

s a

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.

1.x2

#17

x$

5y!

82.

2x2

!2y

2$

3x!

4y#

5pa

rabo

laci

rcle

3.4x

2$

8x#

4y2

$6y

!10

4.8(

x$

x2) #

4(2y

2$

y) $

100

hype

rbol

aci

rcle

5.6y

2$

18 #

24 $

4x2

6.y

#27

x$

y2

ellip

sepa

rabo

la7.

x2#

4(y

$y2

) !2x

$1

8.10

x$

x2$

2y2

#5y

ellip

seel

lipse

9.x

#y2

$5y

!x2

$5

10.1

1x2

$7y

2#

77ci

rcle

hype

rbol

a11

.3x2

!4y

2#

50 !

y212

.y2

#8x

$11

circ

lepa

rabo

la13

.9y2

$99

y#

3(3x

$3x

2 )14

.6x2

$4

#5y

2$

3ci

rcle

hype

rbol

a15

.111

#11

x2!

10y2

16.1

20x2

$11

9y2

!11

8x$

117y

#0

ellip

sehy

perb

ola

17.3

x2#

4y2

!12

18.1

50 $

x2#

120

$y

hype

rbol

apa

rabo

la

Page 66: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A18 Glencoe Algebra 2

Answers (Lesson 8-6)

Skil

ls P

ract

ice

Con

ic S

ectio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

©G

lenc

oe/M

cGra

w-Hi

ll48

7G

lenc

oe A

lgeb

ra 2

Lesson 8-6

Wri

te e

ach

equ

atio

n i

n s

tan

dar

d f

orm

.Sta

te w

het

her

th

e gr

aph

of

the

equ

atio

n i

sa

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.Th

en g

rap

h t

he

equ

atio

n.

1.x2

$25

y2#

25hy

perb

ola

2.9x

2!

4y2

#36

ellip

se3.

x2!

y2$

16 #

0ci

rcle

!%

1#

%1

x2#

y2%

16

4.x2

!8x

!y2

#9

circ

le5.

x2!

2x$

15 #

ypa

rabo

la6.

100x

2!

25y2

#40

0 ellip

se(x

#4)

2#

y2%

25y

%(x

#1)

2!

16#

%1

Wit

hou

t w

riti

ng

the

equ

atio

n i

n s

tan

dar

d f

orm

,sta

te w

het

her

th

e gr

aph

of

each

equ

atio

n i

s a

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.

7.9x

2!

4y2

#36

ellip

se8.

x2!

y2#

25ci

rcle

9.y

#x2

!2x

para

bola

10.y

#2x

2$

4x$

4pa

rabo

la

11.4

y2$

25x2

#10

0hy

perb

ola

12.1

6x2

!y2

#16

ellip

se

13.1

6x2

$4y

2#

64hy

perb

ola

14.5

x2!

5y2

#25

circ

le

15.2

5y2

!9x

2#

225

ellip

se16

.36y

2$

4x2

#14

4hy

perb

ola

17.y

#4x

2$

36x

$14

4pa

rabo

la18

.x2

!y2

$14

4 #

0ci

rcle

19.(

x!

3)2

!(y

$1)

2#

4ci

rcle

20.2

5y2

$50

y!

4x2

#75

ellip

se

21.x

2$

6y2

!9

#0

hype

rbol

a22

.x#

y2!

5y$

6pa

rabo

la

23.(

x!

5)2

!y2

#10

circ

le24

.25x

2!

10y2

$25

0 #

0el

lipse

x

y

O

xy

O4

8

–4 –8 –12

–16

–4–8

x

y

O4

8

8 4 –4 –8

–4–8

y2" 16

x2" 4

x

y

Ox

y

Ox

O

y

48

4 2 –2 –4

–4–8

y2" 9

x2" 4

y2 " 1x2" 25

©G

lenc

oe/M

cGra

w-Hi

ll48

8G

lenc

oe A

lgeb

ra 2

Wri

te e

ach

equ

atio

n i

n s

tan

dar

d f

orm

.Sta

te w

het

her

th

e gr

aph

of

the

equ

atio

n i

sa

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.Th

en g

rap

h t

he

equ

atio

n.

1.y2

#$

3x2.

x2!

y2!

6x#

73.

5x2

$6y

2$

30x

$12

y#

$9

para

bola

circ

lehy

perb

ola

x%

!y

2(x

#3)

2#

y2%

16!

%1

4.19

6y2

#12

25 $

100x

25.

3x2

#9

$3y

2$

6y6.

9x2

!y2

!54

x$

6y#

$81

ellip

seci

rcle

ellip

se#

%1

x2#

(y#

1)2

%4

#%

1

Wit

hou

t w

riti

ng

the

equ

atio

n i

n s

tan

dar

d f

orm

,sta

te w

het

her

th

e gr

aph

of

each

equ

atio

n i

s a

para

bola

,cir

cle,

elli

pse,

or h

yper

bola

.

7.6x

2!

6y2

#36

8.4x

2$

y2#

169.

9x2

!16

y2$

64y

$80

#0

circ

lehy

perb

ola

ellip

se

10.5

x2!

5y2

$45

#0

11.x

2!

2x#

y12

.4y2

$36

x2!

4x $

144

#0

circ

lepa

rabo

lahy

perb

ola

13.A

STR

ON

OM

YA

sat

ellit

e tr

avel

s in

an

hype

rbol

ic o

rbit

.It

reac

hes

the

vert

ex o

f it

s or

bit

at (

5,0)

and

the

n tr

avel

s al

ong

a pa

th t

hat

gets

clo

ser

and

clos

er t

o th

e lin

e y

#x.

Wri

te a

n eq

uati

on t

hat

desc

ribe

s th

e pa

th o

f th

e sa

telli

te if

the

cen

ter

of it

s hy

perb

olic

orbi

t is

at

(0,0

).

!%

1y2" 4

x2" 25

2 " 5x

y

O

x

y

Ox

y

O

(y!

3)2

"9

(x#

3)2

"1

y2" 6.

25x2

" 12.2

5

xO

y

x

y

Ox

y

O

(y #

1)2

"5

(x !

3)2

"6

1 " 3Pra

ctic

e (A

vera

ge)

Con

ic S

ectio

ns

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

Page 67: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A19 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-6)

Rea

din

g t

o L

earn

Math

emati

csC

onic

Sec

tions

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

©G

lenc

oe/M

cGra

w-Hi

ll48

9G

lenc

oe A

lgeb

ra 2

Lesson 8-6

Pre-

Act

ivit

yH

ow c

an y

ou u

se a

fla

shli

ght

to m

ake

con

ic s

ecti

ons?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

6 at

the

top

of

page

449

in y

our

text

book

.

The

fig

ures

in t

he in

trod

ucti

on s

how

how

a p

lane

can

slic

e a

doub

le c

one

tofo

rm t

he c

onic

sec

tion

s.N

ame

the

coni

c se

ctio

n th

at is

for

med

if t

he p

lane

slic

es t

he d

oubl

e co

ne in

eac

h of

the

fol

low

ing

way

s:

•T

he p

lane

is p

aral

lel t

o th

e ba

se o

f th

e do

uble

con

e an

d sl

ices

thr

ough

one

of t

he c

ones

tha

t fo

rm t

he d

oubl

e co

ne.

circ

le•

The

pla

ne is

per

pend

icul

ar t

o th

e ba

se o

f th

e do

uble

con

e an

d sl

ices

thro

ugh

both

of

the

cone

s th

at f

orm

the

dou

ble

cone

.hy

perb

ola

Rea

din

g t

he

Less

on

1.N

ame

the

coni

c se

ctio

n th

at is

the

gra

ph o

f ea

ch o

f th

e fo

llow

ing

equa

tion

s.G

ive

the

coor

dina

tes

of t

he v

erte

x if

the

con

ic s

ecti

on is

a p

arab

ola

and

of t

he c

ente

r if

it is

aci

rcle

,an

ellip

se,o

r a

hype

rbol

a.

a.!

#1

ellip

se;(

3,!

5)

b.x

#$

2(y

!1)

2!

7pa

rabo

la;(

7,!

1)c.

(x$

5)2

$(y

!5)

2#

1hy

perb

ola;

(5,!

5)d.

(x!

6)2

!(y

$2)

2#

1ci

rcle

;(!

6,2)

2.E

ach

of t

he f

ollo

win

g is

the

equ

atio

n of

a c

onic

sec

tion

.For

eac

h eq

uati

on,i

dent

ify

the

valu

es o

f Aan

d C

.The

n,w

itho

ut w

riti

ng t

he e

quat

ion

in s

tand

ard

form

,sta

te w

heth

erth

e gr

aph

of e

ach

equa

tion

is a

par

abol

a,ci

rcle

,ell

ipse

,or

hype

rbol

a.

a.2x

2!

y2$

6x!

8y!

12 #

0A

#;C

#;t

ype

of g

raph

:

b.2x

2!

3x$

2y$

5 #

0A

#;C

#;t

ype

of g

raph

:

c.5x

2!

10x

!5y

2$

20y

!1

#0

A#

;C#

;typ

e of

gra

ph:

d.x2

$y2

!4x

!2y

$5

#0

A#

;C#

;typ

e of

gra

ph:

Hel

pin

g Y

ou

Rem

emb

er

3.W

hat

is a

n ea

sy w

ay t

o re

cogn

ize

that

an

equa

tion

rep

rese

nts

a pa

rabo

la r

athe

r th

anon

e of

the

oth

er c

onic

sec

tion

s?

If th

e eq

uatio

n ha

s an

x2

term

and

yte

rm b

ut n

o y2

term

,the

n th

e gr

aph

is a

par

abol

a.Li

kew

ise,

if th

e eq

uatio

n ha

s a

y2te

rm a

nd x

term

but

no

x2te

rm,t

hen

the

grap

h is

a p

arab

ola.

hype

rbol

a!

11

circ

le5

5pa

rabo

la0

2el

lipse

12

(y!

5)2

"15

(x$

3)2

"36

©G

lenc

oe/M

cGra

w-Hi

ll49

0G

lenc

oe A

lgeb

ra 2

Loci

A l

ocus

(plu

ral,

loci

) is

the

set

of

all p

oint

s,an

d on

ly t

hose

poi

nts,

that

sat

isfy

a gi

ven

set

of c

ondi

tion

s.In

geo

met

ry,f

igur

es o

ften

are

def

ined

as

loci

.For

exam

ple,

a ci

rcle

is t

he lo

cus

of p

oint

s of

a p

lane

tha

t ar

e a

give

n di

stan

cefr

om a

giv

en p

oint

.The

def

init

ion

lead

s na

tura

lly t

o an

equ

atio

n w

hose

gra

phis

the

cur

ve d

escr

ibed

.

Wri

te a

n e

quat

ion

of

the

locu

s of

poi

nts

th

at a

re t

he

sam

e d

ista

nce

fro

m (

3,4)

an

d y

%!

4.

Rec

ogni

zing

tha

t th

e lo

cus

is a

par

abol

a w

ith

focu

s (3

,4) a

nd d

irec

trix

y#

$4,

you

can

find

that

h#

3,k

#0,

and

a#

4 w

here

(h,k

) is

the

vert

ex a

nd 4

uni

tsis

the

dis

tanc

e fr

om t

he v

erte

x to

bot

h th

e fo

cus

and

dire

ctri

x.

Thu

s,an

equ

atio

n fo

r th

e pa

rabo

la is

y#

" 11 6"(x

$3)

2 .

The

pro

blem

als

o m

ay b

e ap

proa

ched

ana

lyti

cally

as

follo

ws:

Let

(x,

y) b

e a

poin

t of

the

locu

s.

The

dis

tanc

e fr

om (

3,4)

to

(x,y

) #th

e di

stan

ce f

rom

y#

$4

to (x

,y).

$(x

$3

#)2

!(

#y

$4)

#2 ##

$(x

$x

#)2

!(

#y

$($

#4)

)2#

(x$

3)2

!y2

$8y

!16

#y2

!8y

!16

(x$

3)2

#16

y

" 11 6"(x

$3)

2#

y

Des

crib

e ea

ch l

ocu

s as

a g

eom

etri

c fi

gure

.Th

en w

rite

an

equ

atio

nfo

r th

e lo

cus.

1.A

ll po

ints

tha

t ar

e th

e sa

me

dist

ance

fro

m (

0,5)

and

(4,

5).

line,

x%

22.

All

poin

ts t

hat

are

4 un

its

from

the

ori

gin.

circ

le,x

2#

y2%

43.

All

poin

ts t

hat

are

the

sam

e di

stan

ce f

rom

($2,

$1)

and

x#

2.

para

bola

,x%

"! 81 "(y

2#

2y#

1)4.

The

locu

s of

poi

nts

such

tha

t th

e su

m o

f th

e di

stan

ces

from

($2,

0) a

nd (

2,0)

is 6

.

ellip

se,"x 92 "

#"y 52 "

%1

5.T

he lo

cus

of p

oint

s su

ch t

hat

the

abso

lute

val

ue o

f th

e d

iffe

renc

e of

the

dis

tanc

es

from

($

3,0)

and

(3,

0) is

2.

hype

rbol

a,"x 12 "

!"y 82 "

%1

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-6

8-6

Exam

ple

Exam

ple

Page 68: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A20 Glencoe Algebra 2

Answers (Lesson 8-7)

Stu

dy

Gu

ide

and I

nte

rven

tion

Solv

ing

Qua

drat

ic S

yste

ms

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

©G

lenc

oe/M

cGra

w-Hi

ll49

1G

lenc

oe A

lgeb

ra 2

Lesson 8-7

Syst

ems

of

Qu

adra

tic

Equ

atio

ns

Lik

e sy

stem

s of

line

ar e

quat

ions

,sys

tem

s of

quad

rati

c eq

uati

ons

can

be s

olve

d by

sub

stit

utio

n an

d el

imin

atio

n.If

the

gra

phs

are

a co

nic

sect

ion

and

a lin

e,th

e sy

stem

will

hav

e 0,

1,or

2 s

olut

ions

.If

the

grap

hs a

re t

wo

coni

cse

ctio

ns,t

he s

yste

m w

ill h

ave

0,1,

2,3,

or 4

sol

utio

ns.

Sol

ve t

he

syst

em o

f eq

uat

ion

s.y

%x2

!2x

!15

x#

y%

!3

Rew

rite

the

sec

ond

equa

tion

as

y#

$x

$3

and

subs

titu

te in

to t

he f

irst

equ

atio

n.

$x

$3

#x2

$2x

$15

0 #

x2$

x$

12Ad

d x

!3

to e

ach

side.

0 #

(x$

4)(x

!3)

Fact

or.

Use

the

Zer

o P

rodu

ct p

rope

rty

to g

etx

#4

orx

#$

3.

Subs

titu

te t

hese

val

ues

for

xin

x!

y#

$3:

4 !

y#

$3

or$

3 !

y#

$3

y#

$7

y#

0

The

sol

utio

ns a

re (

4,$

7) a

nd (

$3,

0).

Fin

d t

he

exac

t so

luti

on(s

) of

eac

h s

yste

m o

f eq

uat

ion

s.

1.y#

x2$

52.

x2!

(y$

5)2

#25

y#x

$3

y#

$x2

(2,!

1),(

!1,

!4)

(0,0

)

3.x2

!(y

$5)

2#

254.

x2!

y2#

9y

#x2

x2!

y#

3

(0,0

),(3

,9),

(!3,

9)(0

,3),

($5!,

!2)

,(!

$5!,

!2)

5.x2

$y2

#1

6.y

#x

$3

x2!

y2#

16x

#y2

$4

",

#, ",!

#,"

,#,

"!,

#, "!

,!#

",

#1

!$

29!"

" 27

!$

29!"

" 2$

30!"

2$

34!"

2$

30!"

2$

34!"

2

1 #

$29!

"" 2

7 #

$29!

"" 2

$30!

"2

$34!

"2

$30!

"2

$34!

"2

Exam

ple

Exam

ple

Exer

cises

Exer

cises

©G

lenc

oe/M

cGra

w-Hi

ll49

2G

lenc

oe A

lgeb

ra 2

Syst

ems

of

Qu

adra

tic

Ineq

ual

itie

sSy

stem

s of

qua

drat

ic in

equa

litie

s ca

n be

sol

ved

by g

raph

ing.

Sol

ve t

he

syst

em o

f in

equ

alit

ies

by g

rap

hin

g.x2

#y2

)25

"x!

#2#

y2*

The

gra

ph o

f x2

!y2

*25

con

sist

s of

all

poin

ts o

n or

insi

de

the

circ

le w

ith

cent

er (

0,0)

and

rad

ius

5.T

he g

raph

of

!x$

"2!

y2+

cons

ists

of

all p

oint

s on

or

outs

ide

the

circ

le w

ith

cent

er !

,0"a

nd r

adiu

s .T

he s

olut

ion

of t

he

syst

em is

the

set

of

poin

ts in

bot

h re

gion

s.

Sol

ve t

he

syst

em o

f in

equ

alit

ies

by g

rap

hin

g.x2

#y2

)25

!&

1

The

gra

ph o

f x2

!y2

*25

con

sist

s of

all

poin

ts o

n or

insi

de

the

circ

le w

ith

cent

er (

0,0)

and

rad

ius

5.T

he g

raph

of

$&

1 ar

e th

e po

ints

“in

side

”bu

t no

t on

the

bra

nche

s of

the

hype

rbol

a sh

own.

The

sol

utio

n of

the

sys

tem

is t

he s

et o

fpo

ints

in b

oth

regi

ons.

Sol

ve e

ach

sys

tem

of

ineq

ual

itie

s be

low

by

grap

hin

g.

1.!

*1

2.x2

!y2

*16

93.

y+

(x$

2)2

y&

x$

2x2

!9y

2+

225

(x!

1)2

!(y

!1)

2*

16 x

y

Ox

y

O6

12

12 6 –6 –12

–6–1

2x

y

O

1 " 2

y2" 4

x2" 16

x2" 9

y2" 4

x2 " 9y2 " 4

x

y

O

5 " 25 " 2

25 " 45 " 2

25 " 45 " 2

x

y

O

Stu

dy

Gu

ide

and I

nte

rven

tion

(c

onti

nued

)

Solv

ing

Qua

drat

ic S

yste

ms

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

Exam

ple1

Exam

ple1

Exam

ple2

Exam

ple2

Exer

cises

Exer

cises

Page 69: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A21 Glencoe Algebra 2

An

swer

s

Answers (Lesson 8-7)

Skil

ls P

ract

ice

Solv

ing

Qua

drat

ic S

yste

ms

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

©G

lenc

oe/M

cGra

w-Hi

ll49

3G

lenc

oe A

lgeb

ra 2

Lesson 8-7

Fin

d t

he

exac

t so

luti

on(s

) of

eac

h s

yste

m o

f eq

uat

ion

s.

1.y

#x

$2

(0,!

2),(

1,!

1)2.

y#

x!

3(!

1,2)

,3.

y#

3x(0

,0)

y#

x2$

2y

#2x

2(1

.5,4

.5)

x#

y2

4.y

#x

( $2!,

$2!)

,5.

x#

$5

(!5,

0)6.

y#

7no

sol

utio

nx2

!y2

#4

( !$

2!,!

$2!)

x2!

y2#

25x2

!y2

#9

7.y

#$

2x!

2(2

,!2)

,8.

x$

y!

1 #

0(1

,2)

9.y

#2

$x

(0,2

),(3

,!1)

y2#

2x"

,1#

y2#

4xy

#x2

$4x

!2

10.y

#x

$1

no s

olut

ion

11.y

#3x

2(0

,0)

12.y

#x2

!1

(!1,

2),

y#

x2y

#$

3x2

y#

$x2

!3

(1,2

)

13.y

#4x

(!1,

!4)

,(1,

4)14

.y#

$1

(0,!

1)15

.4x2

!9y

2#

36(!

3,0)

,4x

2!

y2#

204x

2!

y2#

1x2

$9y

2#

9(3

,0)

16.3

(y!

2)2

$4(

x$

3)2

#12

17.x

2$

4y2

#4

(!2,

0),

18.y

2$

4x2

#4

no

y#

$2x

!2

(0,2

),(3

,!4)

x2!

y2#

4(2

,0)

y#

2xso

lutio

n

Sol

ve e

ach

sys

tem

of

ineq

ual

itie

s by

gra

ph

ing.

19.y

*3x

$2

20.y

*x

21.4

y2!

9x2

'14

4x2

!y2

'16

y+

$2x

2!

4x2

!8y

2'

16

22.G

AR

DEN

ING

An

ellip

tica

l gar

den

bed

has

a pa

th f

rom

poi

nt A

to

poin

t B

.If

the

bed

can

be m

odel

ed b

y th

e eq

uati

on x

2!

3y2

#12

and

the

path

can

be

mod

eled

by

the

line

y#

$x,

wha

t ar

e th

e

coor

dina

tes

of p

oint

s A

and

B?

(!3,

1) a

nd (3

,!1)

1 " 3x

y

B

A

O

x

y

O4

8

8 4 –4 –8

–4–8

x

y

Ox

y

O

1 " 2

©G

lenc

oe/M

cGra

w-Hi

ll49

4G

lenc

oe A

lgeb

ra 2

Fin

d t

he

exac

t so

luti

on(s

) of

eac

h s

yste

m o

f eq

uat

ion

s.

1.(x

$2)

2!

y2#

52.

x#

2(y

!1)

2$

63.

y2$

3x2

#6

4.x2

!2y

2#

1x

$y

#1

x!

y#

3y

#2x

$1

y#

$x

!1

(0,!

1),(

3,2)

(2,1

),(6

.5,!

3.5)

(!1,

!3)

,(5,

9)(1

,0),

",

#5.

4y2

$9x

2#

366.

y#

x2$

37.

x2!

y2#

258.

y2#

10 $

6x2

4x2

$9y

2#

36x2

!y2

#9

4y#

3x4y

2#

40 $

2x2

no s

olut

ion

(0,!

3),( (

$5!,

2)(4

,3),

(!4,

!3)

( 0,(

$10!

)9.

x2!

y2#

2510

.4x2

!9y

2#

3611

.x#

$(y

$3)

2!

212

.$

#1

x#

3y$

52x

2$

9y2

#18

x#

(y$

3)2

!3

x2!

y2#

9

(!5,

0),(

4,3)

((3,

0)no

sol

utio

n((

3,0)

13.2

5x2

!4y

2#

100

14.x

2!

y2#

415

.x2

$y2

#3

x#

$!

#1

y2$

x2#

3

no s

olut

ion

((2,

0)no

sol

utio

n

16.

!#

117

.x!

2y#

318

.x2

!y2

#64

3x2

$y2

#9

x2!

y2#

9x2

$y2

#8

( (2,

($

3!)(3

,0),

"!,

#( (

6,(

2$7!)

Sol

ve e

ach

sys

tem

of

ineq

ual

itie

s by

gra

ph

ing.

19.y

+x2

20.x

2!

y2'

3621

.!

*1

y&

$x

!2

x2!

y2+

16(x

!1)

2!

(y$

2)2

*4

22.G

EOM

ETRY

The

top

of

an ir

on g

ate

is s

hape

d lik

e ha

lf a

n el

lipse

wit

h tw

o co

ngru

ent

segm

ents

fro

m t

he c

ente

r of

the

ellip

se t

o th

e el

lipse

as

show

n.A

ssum

e th

at t

he c

ente

r of

the

ellip

se is

at

(0,0

).If

the

elli

pse

can

be m

odel

ed b

y th

eeq

uati

on x

2!

4y2

#4

for

y+

0 an

d th

e tw

o co

ngru

ent

segm

ents

can

be

mod

eled

by

y#

xan

d y

#$

x,

wha

t ar

e th

e co

ordi

nate

s of

poi

nts

Aan

d B

?

$3#

"2

$3#

"2

BA

(0, 0

)

x

y

O

x

y

O4

8

8 4 –4 –8

–4–8

x

y

O

(x!

2)2

"4

(y$

3)2

"16

12 " 59 " 5

y2" 7

x2" 7

y2" 8

x2" 4

5 " 2

y2" 16

x2" 9

2 " 31 " 3

Pra

ctic

e (A

vera

ge)

Solv

ing

Qua

drat

ic S

yste

ms

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

"!1,

#and

"1,

#$

3!"

2$

3!"

2

Page 70: Chapter 8 Resource Masters - Math Class - Home€¦ · ©Glencoe/McGraw-Hill v Glencoe Algebra 2 Assessment Options The assessment masters in the Chapter 8 Resource Masters offer

© Glencoe/McGraw-Hill A22 Glencoe Algebra 2

Answers (Lesson 8-7)

Rea

din

g t

o L

earn

Math

emati

csSo

lvin

g Q

uadr

atic

Sys

tem

s

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

©G

lenc

oe/M

cGra

w-Hi

ll49

5G

lenc

oe A

lgeb

ra 2

Lesson 8-7

Pre-

Act

ivit

yH

ow d

o sy

stem

s of

equ

atio

ns

app

ly t

o vi

deo

gam

es?

Rea

d th

e in

trod

ucti

on t

o L

esso

n 8-

7 at

the

top

of

page

455

in y

our

text

book

.

The

fig

ure

in y

our

text

book

sho

ws

that

the

spa

cesh

ip h

its

the

circ

ular

for

cefi

eld

in t

wo

poin

ts.I

s it

pos

sibl

e fo

r th

e sp

aces

hip

to h

it t

he f

orce

fie

ld in

eith

er f

ewer

or

mor

e th

an t

wo

poin

ts?

Stat

e al

l pos

sibi

litie

s an

d ex

plai

nho

w t

hese

cou

ld h

appe

n.Sa

mpl

e an

swer

:The

spa

cesh

ip c

ould

hit

the

forc

e fie

ld in

zer

o po

ints

if th

e sp

aces

hip

mis

sed

the

forc

efie

ld a

ll to

geth

er.T

he s

pace

ship

cou

ld a

lso

hit t

he fo

rce

field

in o

ne p

oint

if th

e sp

aces

hip

just

touc

hed

the

edge

of t

hefo

rce

field

.

Rea

din

g t

he

Less

on

1.D

raw

a s

ketc

h to

illu

stra

te e

ach

of t

he f

ollo

win

g po

ssib

iliti

es.

a.a

para

bola

and

a li

ne

b.an

elli

pse

and

a ci

rcle

c.

a hy

perb

ola

and

ath

at in

ters

ect

in

that

inte

rsec

t in

lin

e th

at in

ters

ect

in2

poin

ts4

poin

ts1

poin

t

2.C

onsi

der

the

follo

win

g sy

stem

of

equa

tion

s.

x2#

3 !

y2

2x2

!3y

2#

11

a.W

hat

kind

of

coni

c se

ctio

n is

the

gra

ph o

f th

e fi

rst

equa

tion

?hy

perb

ola

b.W

hat

kind

of

coni

c se

ctio

n is

the

gra

ph o

f th

e se

cond

equ

atio

n?el

lipse

c.B

ased

on

your

ans

wer

s to

par

ts a

and

b,w

hat

are

the

poss

ible

num

bers

of

solu

tion

sth

at t

his

syst

em c

ould

hav

e?0,

1,2,

3,or

4

Hel

pin

g Y

ou

Rem

emb

er

3.Su

ppos

e th

at t

he g

raph

of

a qu

adra

tic

ineq

ualit

y is

a r

egio

n w

hose

bou

ndar

y is

a c

ircl

e.H

ow c

an y

ou r

emem

ber

whe

ther

to

shad

e th

e in

teri

or o

r th

e ex

teri

or o

f th

e ci

rcle

?Sa

mpl

e an

swer

:The

sol

utio

ns o

f an

ineq

ualit

y of

the

form

x2

#y2

'r2

are

all p

oint

s th

at a

re le

ss th

an r

units

from

the

orig

in,s

o th

e gr

aph

isth

e in

terio

rof t

he c

ircle

.The

sol

utio

ns o

f an

ineq

ualit

y of

the

form

x2

#y2

&r2

are

the

poin

ts th

at a

re m

ore

than

run

its fr

om th

e or

igin

,so

the

grap

h is

the

exte

rioro

f the

circ

le.

x

y

Ox

y

Ox

y O

©G

lenc

oe/M

cGra

w-Hi

ll49

6G

lenc

oe A

lgeb

ra 2

Gra

phin

g Q

uadr

atic

Equ

atio

ns w

ith x

y-Te

rms

You

can

use

a gr

aphi

ng c

alcu

lato

r to

exa

min

e gr

aphs

of

quad

rati

c eq

uati

ons

that

con

tain

xy-

term

s.

Use

a g

rap

hin

g ca

lcu

lato

r to

dis

pla

y th

e gr

aph

of

x2#

xy#

y2%

4.

Solv

e th

e eq

uati

on f

or y

in t

erm

s of

xby

usi

ng t

he

quad

rati

c fo

rmul

a.

y2!

xy!

(x2

$4)

#0

To u

se t

he f

orm

ula,

let

a#

1,b

#x,

and

c#

(x2

$4)

.

y#

y#

To g

raph

the

equ

atio

n on

the

gra

phin

g ca

lcul

ator

,ent

er t

he t

wo

equa

tion

s:

y#

and

y#

Use

a g

rap

hin

g ca

lcu

lato

r to

gra

ph

eac

h e

quat

ion

.Sta

te t

he

typ

e of

cu

rve

each

gra

ph

rep

rese

nts

.

1.y2

!xy

#8

2.x2

!y2

$2x

y$

x#

0

hype

rbol

apa

rabo

la

3.x2

$xy

!y2

#15

4.x2

!xy

!y2

#$

9

ellip

segr

aph

is +

5.2x

2$

2xy

$y2

!4x

#20

6.x2

$xy

$2y

2!

2x!

5y$

3 #

0

hype

rbol

atw

o in

ters

ectin

g lin

es

$x

$$

16 $

#3x

2#

""

"2

$x

!$

16 $

#3x

2#

""

"2

$x

($

16 $

#3x

2#

""

"2

$x

($

x2$

4#

(1)(

x2#

$4)

#"

""

2

x

y

O1

–1–2

2

2 1 –1 –2

En

rich

men

t

NAM

E__

____

____

____

____

____

____

____

____

____

____

____

DATE

____

____

____

PERI

OD

____

_

8-7

8-7

Exam

ple

Exam

ple