33
Hyperbola Conic Sections

Hyperbola

  • Upload
    jamar

  • View
    51

  • Download
    1

Embed Size (px)

DESCRIPTION

Hyperbola. Conic Sections. The plane can intersect two nappes of the cone resulting in a hyperbola. Hyperbola. Hyperbola - Definition. A hyperbola is the set of all points in a plane such that the difference in the distances from two points (foci) is constant. - PowerPoint PPT Presentation

Citation preview

Page 1: Hyperbola

Hyperbola

Conic Sections

Page 2: Hyperbola

Hyperbola The plane can intersect

two nappes of the cone resulting in a hyperbola.

Page 3: Hyperbola

Hyperbola - DefinitionA hyperbola is the set of all points in a plane such that the difference in the distances from two points (foci) is constant.

| d1 – d2 | is a constant value.

Page 4: Hyperbola

Finding An Equation

Hyperbola

Page 5: Hyperbola

Hyperbola - DefinitionWhat is the constant value for the difference in the distance from the two foci? Let the two foci be (c, 0) and (-c, 0). The vertices are (a, 0) and (-a, 0).

| d1 – d2 | is the constant.

If the length of d2 is subtracted from the left side of d1, what is the length which remains?

| d1 – d2 | = 2a

Page 6: Hyperbola

Hyperbola - EquationFind the equation by setting the difference in the distance from the two foci equal to 2a.

| d1 – d2 | = 2a

2 2

22

1

2

( )

( )

x c y

d c

d

x y

Page 7: Hyperbola

Hyperbola - EquationSimplify:

2 2 2 2( ) ( ) 2x c y x c y a

Remove the absolute value by using + or -.

2 2 2 2( ) ( ) 2x c y x c y a

Get one square root by itself and square both sides.

2 2 2 2( ) ( ) 2x c y x c y a

2 22 2 2 2( ) ( ) 2x c y x c y a

2 2 2 2 2 2 2( ) ( ) 4 ( ) 4x c y x c y a x c y a

Page 8: Hyperbola

Hyperbola - Equation

Subtract y2 and square the binomials.

2 2 2 2 2 2 2( ) ( ) 4 ( ) 4x c y x c y a x c y a

2 2 2 2 2 2 22 2 4 ( ) 4x xc c x xc c a x c y a

Solve for the square root and square both sides.

2 2 24 4 4 ( )xc a a x c y

222 2 2( )xc a a x c y

2 2 2( )xc a a x c y

Page 9: Hyperbola

Hyperbola - Equation

Square the binomials and simplify.

222 2 2( )xc a a x c y

2 2 2 4 2 2 22 ( )x c xca a a x c y

2 2 2 4 2 2 2 22 2x c xca a a x xc c y 2 2 2 4 2 2 2 2 2 2 22 2x c xca a a x xca a c a y 2 2 4 2 2 2 2 2 2x c a a x a c a y

Get x’s and y’s together on one side. 2 2 2 2 2 2 2 2 4x c a x a y a c a

Page 10: Hyperbola

Hyperbola - Equation

Factor.

2 2 2 2 2 2 2 2 4x c a x a y a c a

2 2 2 2 2 2 2 2x c a a y a c a

Divide both sides by a2(c2 – a2)

2 2 2 2 2 22 2

2 2 2 2 2 2 2 2 2

x c a a c aa y

a c a a c a a c a

2 2

2 2 21

x y

a c a

Page 11: Hyperbola

Hyperbola - Equation

Let b2 = c2 – a2

2 2

2 2 21

x y

a c a

2 2

2 21

x y

a b where c2 = a2 + b2

If the graph is shifted over h units and up k units, the equation of the hyperbola is:

Page 12: Hyperbola

Hyperbola - Equation

2 2

2 21

x h y k

a b

where c2 = a2 + b2

Recognition:How do you tell a hyperbola from an ellipse?

Answer:A hyperbola has a minus (-) between the terms while an ellipse has a plus (+).

Page 13: Hyperbola

Graph - Example #1

Hyperbola

Page 14: Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Center: (-1, -2)

The hyperbola opens in the “x” direction because “x” is positive.

Transverse Axis: y = -2

Page 15: Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Vertices (2, -2) (-4, -2)

Construct a rectangle by moving 4 units up and down from the vertices.

Construct the diagonals of the rectangle.

Page 16: Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Draw the hyperbola touching the vertices and approaching the asymptotes.

Where are the foci?

Page 17: Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

The foci are 5 units from the center on the transverse axis.

2 2 2c a b 2

2

9 16

25

5

c

c

c

Foci: (-6, -2) (4, -2)

Page 18: Hyperbola

Hyperbola - Graph

2 21 2

19 16

x y

Graph:

Find the equation of the asymptote lines.

42 13y x

Slope =

Use point-slope formy – y1 = m(x – x1) since the center is on both lines.

3

4

-4

4

3

Asymptote Equations

Page 19: Hyperbola

Graph - Example #2

Hyperbola

Page 20: Hyperbola

Hyperbola - Graph

2 210 5 40 10 15 0y x y x

Sketch the graph without a grapher:

Recognition:How do you determine the type of conic section?

Answer:The squared terms have opposite signs.

Write the equation in hyperbolic form.

Page 21: Hyperbola

Hyperbola - Graph

2 210 5 40 10 15 0y x y x

Sketch the graph without a grapher:

2 210 40 5 10 15y y x x

2 210 4 ?? 5 2 ?? 15y y x x

2 210 4 5 2 54 1 401 5y y x x

2 210 2 5 1 50y x

2 210 2 5 1 50

50 50 50

y x

Page 22: Hyperbola

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Center: (-1, 2)

Transverse Axis Direction:Up/Down

Equation: x=-1

Vertices: Up/Down from the center or 5 1, 2 5

Page 23: Hyperbola

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Plot the rectangular points and draw the asymptotes.

Sketch the hyperbola.

Page 24: Hyperbola

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Plot the foci.

Foci:

2 2 2

2

2

10 5

15

15

c a b

c

c

c

1, 2 15

Page 25: Hyperbola

Hyperbola - GraphSketch the graph

without a grapher:

2 22 1

15 10

y x

Equation of the asymptotes:

52 1

10y x

22 1

2y x

Page 26: Hyperbola

Finding an EquationA problem for CSI!

Hyperbola

Page 27: Hyperbola

Hyperbola – Find an EquationThe sound of a gunshot was recorded at one microphone 0.54 seconds before being recorded at a second microphone. If the two microphones are 2,000 ft apart. Provide a model for the possible locations of the gunshot. (The speed of sound is 1100 ft/sec.)

The time between the shots can be used to calculate the difference in the distance from the two microphones.

1100 ft/sec * 0.54 sec = 594 ft. The constant difference in distance from the microphones is 594 ft.

Since the difference is constant, the equation must be a hyperbola. The points on the hyperbola are possible positions for the gunshot.

Page 28: Hyperbola

Hyperbola – Find an EquationTwo microphones are stationed 2,000 ft apart. The difference in distance between the microphones is 594 ft.

Let the center be at (0,0). The foci must be 2,000 ft apart.

The vertices are a possible position for the gunshot. The difference in the distance must be 594 feet between the vertices.

V

Let the vertices be at (+z, 0). Assuming z>0, then(z-(-1000)) – (1000-z) = 594z+1000-1000+z = 5942z = 594 or z = 297.

Page 29: Hyperbola

Hyperbola – Find an Equation

V

Start finding the model of the hyperbola.

V(297, 0)

V

V(-2970, 0)

2 2

21

88209

x y

b

The distance from the center to the foci (c) is 1000 ft. Find b.

Oops! We could have remembered the constant difference in distance is 2a! 2a = 594, a = 297.

2972 = 88209

Page 30: Hyperbola

Hyperbola – Find an Equation

V

V(294, 0)

V

V(294, 0)

The model is:

2 2 2

2 2 2

2

1000 297

911791

c a b

b

b

2 2 2 2

2 21 1

88209 911791 297 954.9

x y x yor

Page 31: Hyperbola

Hyperbola – Find an EquationThe gunshot was calculated to be at some point along the hyperbola.

Page 32: Hyperbola

Conic Section Recogition

Page 33: Hyperbola

Recognizing a Conic SectionParabola -

One squared term. Solve for the term which is not squared. Complete the square on the squared term.

Ellipse - Two squared terms. Both terms are the same “sign”.

Circle - Two squared terms with the same coefficient.

Hyperbola -

Two squared terms with opposite “signs”.