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Supplementary Modules for Pre-Service Mathematics Teachers

Guershon Harel, Jeff Rabin, Laura Stevens, Evan Fuller

University of California, San Diego

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1. Conic Sections A mathematical topic has a physical/perceptual aspect, a geometric description, and an

algebraic formulation. Important features visible in any of these three aspects should be visible in the others also, and one should search for and exploit these correspondences. The organizing principle of this module is the interrelationships between physical/perceptual, geometric, and algebraic forms. We refer to this principle as the PGA way of thinking. Throughout the module, we are developing the PGA Way of Thinking, which is valuable and used by mathematicians. A central instructional objective is to help students acquire this Way of Thinking. The specific context (circles, ellipses, etc) chosen to implement this objective is only a vehicle, and less important than the objective itself. The particular context of conic sections is very rich and, in our view, suitable for high school students.

Beginning with the circle as the most familiar conic section, and continuing through the others, we follow the sequence: perception, geometry, algebra. That is, what is our intuitive understanding of circle based on our perception of physical examples? Then the need for communicationin this case the need to communicate to others a precise description of the curve formalizing the intuitiondemands a characterization of the curve as a geometric locus. The adequacy of the geometric description is tested by formally proving some intuitive properties about the curve. Together with the need for communication, we utilize the need for computation. Some problems about these properties invoke computation, in the form of algebraic representations of the curve in terms of Cartesian or parametric equations and manipulations of these representations. With geometric and algebraic descriptions in hand, new properties of the locus can be discovered and proved. Proving, in general, draws on the need for certaintyto know that something is true. Truth alone, however, is not our only aim, and we desire to educate students to strive to know why something is truethe cause that makes it truea need we refer to as the need for causality. Finally, by comparing and reflecting on the results, we aim to instill in the students the desire to reveal a structurethe need for structure. Specifically, our aim is to lead students to recognize a common structure among all the conic sectionsgeometric as well as algebraic. This sequence from perception through formalization and generalization characterizes the development of many mathematical concepts. It serves to humanize mathematics, showing students that definitions and theorems are not handed down by the gods but arise from human experience. We have designed problems giving students ample opportunity to develop and explore all these viewpoints (physical/perceptual, geometric, and algebraic) and to contrast perceptual justifications with geometric and algebraic proofs of the properties of the conic sections. We want students to understand that the mathematical definition of a circle, say, is the basis for deriving all the properties of this locus, and the means for proving that some newly encountered object is or is not a circle. Although we begin with perception, it is critical that students also encounter situations in which perception alone leads to ambiguous or incorrect expectations. The need for certainty compels students to resolve these situations by deductive reasoning, which, in turn, promotes the replacement of perceptual reasoning by deductive reasoning.

These five needs manifest a crucial principle, called the necessity principle. It claims: For students to learn what we intend to teach them, they must have a need for it, where need refers to intellectual need, not only psychological need. Intellectual need has to do with disciplinary knowledge being born out of peoples current knowledge through engagement in problematic situations conceived as such by them. Psychological need, on the other hand, has to do with peoples desire, volition, interest, self determination, and the like. Indeed, before one immerses

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oneself in a problem, one must be willing to engage in the problem and persist in the engagement. Our focus in this module is on intellectual rather than psychological needs. As the module unfolds, we urge the reader to contrast this Necessity approach with the current standards-driven approach in high school teaching.

The module also emphasizes another crucial principle: the repeated reasoning principle. It claims: Students must practice reasoning in order to internalize, organize, and retain the mathematics they have learned. Repeated reasoning, not mere drill and practice of routine problems, is essential to the process of internalizationa state where one is able to apply knowledge autonomously and spontaneously. The sequence of problems must continually call for reasoning through the situations and solutions and must respond to the students changing intellectual needs.

As the discussion of each curve moves from visual intuition to geometric characterization to algebraic formalization, natural questions recur in each case, for example, from does a circle uniquely determine its center? to does an ellipse uniquely determine its foci? Students should develop a structural way of thinking, where they spontaneously ask these questions and explore the relationships between properties of different conic sections. The Dandelin sphere construction will be presented for the ellipse, but students must rethink it carefully for the parabola and hyperbola. Since the focus/directrix definition applies uniformly to all conics, students must carefully formulate the reasons for their differing appearances. The problem of finding tangent lines recurs for each curve. Algebraic techniques such as completing the square also apply to the equations of all conics and must be reasoned out in each case. Repeated reasoning, in the context of the PGA way of thinking, promotes in particular attention to meaning, especially for the algebraic symbols, which we call referential symbolic reasoning. Since the algebraic symbols can have both geometric and intuitive meanings, students have numerous opportunities to interpret their algebraic manipulations in geometric or physical terms. As a result they notice patterns and opportunities to simplify calculations which would otherwise be overlooked. A related goal is for students to develop the algebraic invariance way of thinkinga habit of mind where one manipulates an algebraic expression not haphazardly but with the purpose of arriving at a desired form and maintaining certain properties of the expression invariant.

In designing and teaching this module, we have struck balances in emphasis between several aspects of the subject which are in tension. These tensions include:

Specific properties of each conic versus general properties common to all. The use of elementary methods suitable for high school presentation versus links to

more advanced (calculus, linear algebra) methods. Use of synthetic versus analytic geometry methods. Emphasis on 3d (sections of a cone) versus 2d (focus/directrix) definitions of the

conics. Lastly, our approach is to carefully attend to subject matterdefinitions, theorems,

proofs, problems and their solutions, etc.as well as to ways of thinking (WoT), such the PGA way of thinking, referential symbolic way of thinking, and algebraic invariance mentioned above. We will refer to elements of subject matter as ways of understanding, to differentiate them from ways of thinking. For example, the following are different ways of understanding the phrase derivative of a function at a , or the symbol ( )f a : the slope of a line tangent to the graph of a function at a or the

0lim ( ) ( ) /h

f a h f a h or the instantaneous rate of change

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at a or the slope of the best linear approximation to a function near a . Other ways of understanding and ways of thinking will emerge as the module unfolds.

The module consists of six units. Each unit begins with a list of focus ways of understanding and way of thinking, and proceeds with the classroom problems that attend to them. A pedagogical discussion on these problems, including observations from our own classes, then follows. The unit concludes with a set of practice problems.

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1.1 The Circle

1.1.1 Focus Ways of Thinking and Ways of Understanding PGA way of thinking: Attending to interrelationships between physical/perceptual,

geometric, and algebraic realities. Algebraic invariance way of thinking. An equation can be rewritten in various forms

which make certain properties more noticeable. The link between the forms is provided by something which remains invariant, for example the solution set.

Algebraic way of thinking. This is a broad way of thinking. One of its instantiations is the realization that when applying algebra, to a geometry problem for example, one must tell algebra all the relevant geometric constraints. Likewise, we may talk of a geometric way of thinking in this manner; namely, in solving a geometry problem, one must tell geometry all the given conditions.

Usefulness of completing the square. Understanding the concept of tangent line as a line intersecting a circle only once.

Understanding a property derived from this definition: the tangent line at a point is perpendicular to the radius to that point.

We begin with an object familiar to students both perceptually and mathematically: the circle. We draw a circle and ask, what makes this a circle? How do we recognize a circle or communicate it to someone else? We hope to elicit the characterization as a locus: the set of points in a plane equidistant from a specified point. This could be communicated by specifying the center and radius (and the plane, if not obvious). This definition translates easily into the usual Cartesian equation for a circle centered at the origin, or at any other point. The following problems may be used in class to solidify these ideas through the necessity principle and the repeated reasoning principle.

Classroom Problem 1: You drew a circle with a compass but forgot to mark its center. It is possible to retrieve the center of the circle? Classroom Problem 2: Some of the solutions to the previous problem led us to conclude that an equation of the form 2 2 2( ) ( )x a y b r represents a circle with center ( , )a b and radius r . Write down your meaning of the statement a circle with center ( , )a b and radius r is represented by the equation 2 2 2( ) ( )x a y b r . Explain again why this is the case. Now consider the circle 2 2( 1) ( 2) 3x y . Tami expanded this equation into the equation 2 2 2 4 2 0.x y x y Bruce didnt see the original equation. Can Bruce retrieve the center and radius of the circle from the second equation? Classroom Problem 3: In the previous two problems, you learned that a circle with center ( , )a b and radius r has the equation 2 2 2( ) ( )x a y b r . The problem that Bruce encountered can be generalized: If we expand the brackets in the circles equation and collect the corresponding terms, we can rewrite the above equation in the form 2 2 2 2 22 2 ( ) 0x y ax by a b r . This shows that the equation of a circle can be written in two different forms: as

1. 2 2 2( ) ( )x a y b r or as

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2. 2 2 2 2 22 2 ( ) 0x y ax by a b r . We can make the second equation more compact and easier to remember by substituting 2f a , 2g b , and 2 2 2h a b r to obtain

2 2 0.x y fx gy h This shows that every circle can be represented by an equation of the form

3. 2 2 0.x y fx gy h Question: Does every equation of the form 2 2 0x y fx gy h represent a circle? Classroom Problem 4: You are familiar with the concept angle between two intersecting lines.

1. How would you define angle between two intersecting planes? 2. How would you define angle between a line and plane (if they intersect)? 3. How would you define angle between two intersecting curves? 4. We often talk about orthogonal lines, orthogonal planes, and a line

orthogonal to a plane. Can we talk about orthogonal curves? Classroom Problem 5: Construct two orthogonal circles in one plane. Classroom Problem 6: A circle of radius 4 is centered at (1,2) . Where does it cross the coordinate axes? What lines are tangent to it at these points? Classroom Problem 7: Prove that the intersecting circles 2 2 1 1 1 0x y f x g y h

and 2 2 2 2 2 0x y f x g y h are orthogonal if and only if 1 2 1 2 1 22( ).f f g g h h 1.1.2 Pedagogical Considerations

A recurring theme of this unit (and, indeed, all the units of this module) is that both geometry and algebra are systems for drawing logical conclusions from given data. To exploit the power of these systems by drawing the strongest conclusions, it is necessary to tell geometry or tell algebra all the

given conditions: the conditions must be stated in a form which these systems can process, and all must be used nontrivially in the reasoning. This is what we referred to earlier as the algebraic way of thinking and geometric way of thinking.

A geometric solution to this problem begins by choosing three points 1P , 2P , 3P on the circle. Because three (noncollinear) points determine a circle, these three points suffice to tell geometry which circle is given. To tell geometry that we want the circle through these three points, rather than some other curve, we use the locus definition of the circle: the three points are equidistant from the center. The locus of points equidistant from any two of the points is the perpendicular bisector of the line segment joining them. Each of these bisectors therefore passes through the center, which can be found as the intersection of any two of them. This solution is obviously related to the theorem that the perpendicular bisectors of the sides of any triangle are concurrent, in fact meeting at the center of the circumscribed circle. Traditionally, the center of a circle is found by simply taking two non-parallel chords of the circle and finding the intersection of their perpendicular bisectors. The two solutions are mathematically the same, of course. However, the former solution is much more explicit about the reasoning process underlying the

Classroom Problem 1: You drew a circle with a compass but forgot to mark its center. It is possible to retrieve the center of the circle?

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constructionthat there was a reason to think of the two chords. In our approach to this module, we aim at demystifying mathematics for the students, in that solutions to mathematical problems do not emerge from nothing but from careful reasoning and a representation of the necessary and sufficient conditions of the problem.

There is an algebraic solution mirroring this geometric one. Let the point iP have coordinates ( , )i ix y . The conditions telling algebra that the three points lie on a circle with center ( , )a b and radius r are the three equations 2 2 2( ) ( )i ix a y b r . They state that each point is r units from the center. Subtracting a pair of these equations gives a condition that the center is equidistant from two of the given points, the specific distance r having cancelled out (referential

symbolic reasoning). This turns out to be, for example, 1 2 1 2 1 22 12 2

y y x x x xb ay y

, precisely the statement that ( , )a b lies on the perpendicular bisector of the segment 1 2PP . This is the algebraic proof of the claim that the perpendicular bisector is the locus of equidistant points. The center can be found by solving any two of these linear equations simultaneously. The third linear equation is dependent on the two chosen, and therefore the center lies on it too.

Students may wish to locate the center as the intersection of two diameters. How easy this is depends on the available tools. Given a ruler, one might locate two points on the circle a maximal distance apart and join them. (Is this an allowed use of a ruler?) With straightedge and compass, it can be done by inscribing a right angle at a point of the circle; its sides will meet the circle at ends of a diameter. This could necessitate the theorem that an inscribed angle measures half of the intercepted arc. It is useful for students to realize how geometric solutions depend on the available tools.

In our classroom one student drew a circle on a sheet of paper and folded the paper so as to bring one half of the circle onto the other, thus making the fold line a diameter. Two such diameters intersect at the center. This is an excellent example of an empirical/perceptual proof as opposed to a deductive one. The perceptual approach is valuable for highlighting intuitive properties of the circle, which should be deductively verifiable later: its symmetry about any diameter, for example.

The first part of this problem is to ensure that students understand the meaning of the idea an equation represents a plane curve: that every ordered pair satisfying the equation is a point on the curve, and conversely, the ordered pair of every point on the curve satisfies the equation. Our experience suggests that this meaning is not always clear to students. This issue goes beyond the phenomenon that students are not careful to distinguish a statement from its converse, sometimes assuming that a chain of reasoning proves an if and only if claim even though the reasoning is not reversible. Having an equation for some locus is so powerful that students are unlikely to question the correspondence between the locus and the equationthey see deriving an

Classroom Problem 2: Some of the solutions to the previous problem led us to conclude that an equation of the form

2 2 2( ) ( )x a y b r represents a circle with center ( , )a b and radius r . Write down your meaning of the statement a circle with center ( , )a b and radius r is represented by the equation 2 2 2( ) ( )x a y b r . Explain again why this is the case. Now consider the circle

2 2( 1) ( 2) 3x y . Tami expanded this equation into the equation 2 2 2 4 2 0.x y x y Bruce didnt see the original equation. Can Bruce retrieve the center and radius of the circle from the second equation?

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equation as the end, without separating the steps of showing that (1) every point in the locus satisfies the equation and (2) showing the every point that satisfies the equation is in the locus. To help students understand this issue, we gave it repeated attention when working with the different conic sections.

Students worked on the second part of the problem in groups and reconstructed the original form of the circle's equation by completing the square. This promotes the Algebraic Invariance way of thinking and introduces the usefulness of completing the square, an essential technique throughout this module (Repeated Reasoning). Again students completed the square to recast the equation as

2 2 2 2

2 2 4f g f gx y h . This

represents a circle iff the constant on the right side is a positive number 2r , that is iff 2 2 4f g h . Students should ask (or be asked) what is the locus if this condition is not met: a single point, or else empty.

For the angle between two planes, students suggested choosing a point in the intersection of the planes, choosing a line through this point in each of the planes, and taking the angle between these lines. They realized that this concept is not well-defined, because it depends on the choices made. They decided instead to use the angle between the planes normal lines, which is not ambiguous. This productive disequilibrium is a very important experience in terms of the origin of mathematical definitions. First, new concepts often rely on old ones: the angle between planes is defined in terms of the already-understood angle between lines. However, if a definition allows different people to make different choices, which lead to different results, then the Need for Communication requires

that these choices be standardized in some way so that everyone obtains the same answer. Similarly, the angle between a line and a plane cannot be defined as simply the angle

Classroom Problem 4: You are familiar with the concept angle between two intersecting lines. 1. How would you define angle

between two intersecting planes? 2. How would you define angle

between a line and plane (if they intersect)?

3. How would you define angle between two intersecting curves?

We often talk about orthogonal lines, orthogonal planes, and a line orthogonal to a plane. Can we talk about orthogonal curves?

Classroom Problem 3: In the previous two problems, you learned that a circle with center ( , )a b and radius r has the equation

2 2 2( ) ( )x a y b r . The problem that Bruce encountered can be generalized: If we expand the brackets in the circles equation and collect the corresponding terms, we can rewrite the above equation in the form 2 2 2 2 22 2 ( ) 0x y ax by a b r . This shows that the equation of a circle can be written in two different forms: as 1. 2 2 2( ) ( )x a y b r or as 2. 2 2 2 2 22 2 ( ) 0x y ax by a b r . We can make the second equation more compact and easier to remember by substituting

2f a , 2g b , and 2 2 2h a b r to obtain 2 2 0.x y fx gy h This shows that every circle can be represented by an equation of the form 3. 2 2 0.x y fx gy h Question: Does every equation of the form

2 2 0x y fx gy h represent a circle?

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between the line and any chosen line in the plane which meets it. The angle between the line and its orthogonal projection into the plane will do, and this is equivalent to taking the complement of the angle between the line and the planes normal.

To define the angle between curves, and specifically the idea of orthogonal curves, students suggested using their tangent lines. Asked why this makes sense, they responded that the lines locally approximate the curves and we already know what the angle between lines means. This is a proper answer. The definition for the angle between two curves emerged naturally for the students from their image of tangent line as a local approximation to a curve. Here the instructor can discuss the relationship between this meaning of tangent line and Euclids meaning as a line intersecting exactly once (which may be familiar to students from high-school). This is part of the need for communication, in that different meanings of the same term must be equivalent, for otherwise it wouldnt be possible to communicate meaningfully about the term. Furthermore, this would advance another important way of thinkingthat a term can have multiple interpretations, and it is advantageous to have multiple interpretations for a term. Once this equivalency has been established, the instructor may ask, Imagine that I am blind. How can I prove that given a circle, there is a line that does not intersect it? The game of playing blind is to remove the use of visual perception, which is common among students. A geometric approach to this depends on a careful axiomatization of geometry, so algebra is more appropriate here. If the circle and line have equations 2 2 2( ) ( )x a y b r and y mx k , how many intersection points can they have? Students should explain clearly why substituting y mx k into the equation of the circle has the meaning of locating intersection points. Since a

quadratic equation results, there are two, one, or no intersection points, and any of these possibilities can occur by adjusting the parameters. The case with no intersections answers the given question, and the case of a single intersection defines the idea of a tangent line.

In our classroom, a student presented her approach (see Figure 1). Circles centered at O and O intersect at A and B ; lines OO and AB are drawn and meet at C . She suggested that the radii must be greater than half OO in order for the circles to intersect. Such conjectures should always be examined carefully, using perceptual evidence, relevant theorems, potential counterexamples, and so forth, so that students progressively develop their skepticism and standards of proof. The instructor asked what if the radii are much greater than OO ? Homework: Determine the necessary and sufficient condition on the radii for the circles to meet. The student made several further conjectures, which the class examined critically: All four triangles such as OAC are congruent (in

fact there are two congruent pairs), OO AB and bisects AB ; the circles will be orthogonal if their radii are equal.

A second student presented this reasoning. Calling the radii 1 2,r r and the distance

Classroom Problem 5: Construct two orthogonal circles in one plane.

C

B

A

O O'

Figure1:Constructingtwoorthogonalcircles

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between the centers s , the circles will be orthogonal if each radius to A or B is tangent to the other circle, that is, if the radii are perpendicular there. This happens if 2 2 21 2r r s (by the converse to the Pythagorean theorem). Thus, the construction can be made as follows. Draw one circle with any center O and radius 1r ; at the endpoint A of a radius, draw a perpendicular segment AO of any length 2r , and a second circle centered at O will be orthogonal to the first circle at A . Why also at the second intersection B ? By the earlier argument that there are pairs of congruent triangles! So students see that the first approach was not wrong or a waste of time; the reasoning is valuable after we see how it fits into the structure of the problem.

The equation of the circle is 2 2( 1) ( 2) 16x y , from which the intercepts are easily found as (0,2 15) and (1 2 3,0) . They also follow from applying the Pythagorean theorem to a diagram, which should not be surprising since the Pythagorean theorem is the basis for the equation of the circle.

They are of course symmetric about the vertical and horizontal diameters of the circle. To find the tangent lines, we need to tell algebra or geometry what a tangent line is. As

we saw, for circles one can use the simplest definition: a line meeting the circle in exactly one point. However, students should have no trouble giving examples of lines which meet a parabola in one point but are not tangent to it, or lines tangent to a non-convex closed curve which meet it more than once.

As Euclid shows, the line perpendicular to the radius of a circle at some point P on the circle meets the circle only there and is otherwise external to it. Indeed, if O is the center and Q some other point on this line, then OQ is greater than the radius OP by the Pythagorean theorem, so Q is outside the circle. This can be confirmed by students who know calculus: implicit differentiation of the equation of the circle gives / ( 1) / ( 2)dy dx x y , the negative reciprocal of the slope of the radius to ( , )x y . The slopes are 1 / 15 at the y -intercepts and

3 at the x -intercepts, which look plausible from a picture. The two signs again reflect the symmetry of the circle about its diameters. Since the equation of the circle presents it as a level set of a function ( , )f x y , one can also find a normal vector by taking the gradient, which gives twice the radius vector. One can finesse the meaning of the tangent line by simply defining it to be the line having the slope computed by calculus, but this gives no insight unless the process of differentiation is understood.

An algebraic approach may lead to more insight than calculus for many students. We seek a line y mx b intersecting the circle only once, say at a y-intercept 0x . Solving the equations of the line and circle simultaneously leads to the quadratic

2 2 2(1 ) 2( 2 1) ( 4 11) 0.m x mb m x b b If this has only one root 0x , then two coefficients must vanish: 2 4 11 0b b gives 2 15b , so that the line and the circle have the same y-intercept, and then 1 / ( 2) 1 / 15m b , the expected slope.

Classroom Problem 6: A circle of radius 4 is centered at (1,2) . Where does it cross the coordinate axes? What lines are tangent to it at those points?

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Orthogonality condition for circles. This is a beautiful example of the PGA WoT (way of thinking): a natural geometric condition, orthogonality of circles, is equivalent to a simple algebraic condition on the coefficients in their equations. This is also an opportunity for students to realize that equivalence (if and only if) means there are two distinct implications to prove, an idea that will recur (Repeated Reasoning!) often. If the circles are orthogonal,

with radii 1 2,r r and centers separated by s , then the Pythagorean Theorem implies 2 2 2

1 2r r s . Completing the squares in the equations of the circles, we find their radii given by

2 2 21 1 1 1

1 ( )4

r f g h , 2 2 22 2 2 21 ( )4r f g h . Their centers are 1 1( / 2, / 2)f g and 2 2( / 2, / 2)f g ,

so the distance formula gives 2 2 2 2 21 1 2 2 1 1 2 21 ( 2 2 ).4

s f f f f g g g g Substituting indeed produces the claimed algebraic condition. Conversely, if the condition holds we can reverse the reasoning and deduce 2 2 21 2r r s , whereupon the converse to the Pythagorean Theorem shows that the circles are orthogonal. Students may not realize that the converse to the Pythagorean Theorem is distinct from the theorem itself and requires a distinct proof. It is useful to keep track of the results from geometry that are being used to support the reasoning throughout this module, and students should independently make sure they can prove them. Of course, there are calculus-based approaches to this problem using the slopes of the tangent (or normal) lines to verify orthogonality.

A final question for students, to clarify the subtle logic in this problem: Does the given algebraic condition ensure that the circles intersect, or does it only make them orthogonal assuming they intersect? Since the condition implies 2 2 21 2r r s , if follows that there is a right triangle having sides 1 2, ,r r and s . Then the triangle inequality says 1 2r r s and 1 2| |r r s , which means that the centers of the circles are close enough for them to intersect. So the condition does guarantee that the circles intersect, and orthogonally. However, it does not guarantee that the squared radii computed from the equations are actually positive; this has to be assumed.

1.1.3 Supplementary and Practice Problems The following problems should be seen as a continuation of the Classroom Problems.

Problem 1. Choose any values for f , g , and h in the equation 2 2 0x y fx gy h and convert the resulting equation into an equation of the

form 2 2 2( ) ( )x a y b r . Now choose another set of values for f , g , and h , but this time f should be a fraction, g an irrational number, and h a negative integer. Again convert the resulting equation into an equation of the form

2 2 2( ) ( ) .x a y b r Problem 2. Find the equation of a circle centered at the origin that is tangent to the line 2 2 39x y . Problem 3: Let P, Q, R, and S be four distinct points in the plane such that no three

Classroom Problem 7: Prove that the intersecting circles

2 21 1 1 0x y f x g y h

and

2 22 2 2 0x y f x g y h are

orthogonal if and only if 1 2 1 2 1 22( ).f f g g h h

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of them are collinear. a. How many circles can you draw through the point P? b. How many circles can you draw through the points P and Q? c. How many circles can you draw through the points P, Q and R? d. How many circles can you draw through the points P, Q, R and S?

Problem 4: Find the intersection points of the circles 2 2( 1) ( 1) 5x y and 2 2( 1) 4x y . Are the circles orthogonal at these points?

Problem 5: Find all the common tangents to the circles 2 2( 2) 1x y and 2 2( 2) 1.x y

Problem 6: How many common tangents can there be for two circles in the plane? Exhibit all the possibilities, and prove that there are no others. Problem 7: Given 2 intersecting circles in a plane, consider the 4 tangents at the points of intersection (1 tangent at each of 2 points for each of 2 circles). Draw perpendicular lines from each circle center to each tangent, thus obtaining 6 points of intersection: 4 lines from each center yields 8 points, only 6 of which are distinct. Show that all these points (A,B,C,D,E,F) are on the same circle.

Problem 8: We have seen that an equation of the form 2 2 2( ) ( )x a y b r represents a circle in the plane with center ( , )a b and radius r .

a. Define sphere geometrically. b. What is the equation of a sphere with center ( , , )a b c and radius r ? c. Given two equations of two spheres, find an algebraic condition which

ensures the two spheres (i) do not intersect, (ii) intersect at exactly one point, (iii) intersect at more than one point.

d. For the last case, where the two spheres intersect at more than one point, what

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type of locus is the intersection of the two spheres? Problem 9: For what values of p will the following system have exactly three solutions?

2 2 9

( )( 3) 0

x y

py x x p

In our classroom we did not introduce parametric equations of the circle. Some of the

following problems serve to do so. In Problem 8, for example, students need to use the definition of a circle, including the part about being planar, to verify that the given curves are circles. They also need to observe that parametric equations describe not only a locus, but also the motion of a point along that locus in time. This motion is useful for physical modeling but irrelevant for geometric properties of the locus. The meaning of the parameter t needs to be deeply understood in order to distinguish between intersection of the loci and collision of the points moving along them. Here algebra encodes more information than the geometry requires.

Problem 10: A particle moves along the line 32 2 05

x y . At 12:00 AM, the

particles x -coordinate is 22

. What is the particles y -coordinate at that time? The

particles velocity in the direction of the x -axis is 49

meter/second. What is the particles position (in the x - y plane) at any given time t ? What is the particles velocity along the y -axis? Problem 11: The position function ( ) ( ( ), ( ), ( ))t x t y t z tr is often written as

( ) ( ) ( ) ( )t x t y t z t i j kr . Consider the two parametric curves: 1( ) (cos ) (sin )t t t r i j and 2( ) (cos ) (sin )t t t r j k . What are these curves? Do they

intersect? Problem 12: A particle moves so that its position in the plane at time t seconds from the start of its movement is (cos ,sin )t t . If the particles movement was observed for 2 seconds, what curve did the particle complete? If the particles movement was observed just for

2 seconds, what curve did the particle complete? Answer the same

question for 4 seconds and for 1 second. Problem 13: Three particles, A, B, and C, move in space. The positions of the three objects in space at time t seconds from the start of their movement are, respectively:

(cos ,sin 2 ,0)t t , (cos , 0,sin )2t t , (0,cos2 ,sin 4 )t t . What are the curves along which

the particles move? Can you tell which of the particles is the fastest or the slowest? Problem 14: The hands of a clock have lengths 4 and 8 inches. Where are they located t minutes after noon? What is the angle between them at that time, and what is the distance between their tips? At what times do the hands coincide?

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2. Unit 2: The Ellipse Focus Ways of Thinking and Ways of Understanding

Referential symbolic way of thinking. Attending, when there is a need, to the meaning of symbols and their manipulations.

Deductive reasoning: Logical structure of proofs: what is given, what is proved, what can be assumed or chosen freely. Distinguishing between a theorem and its converse.

Algebra as deduction: Every algebraic computation is a proof of something. Try to state explicitly what it proved.

PGA way of thinking: Attending to interrelationships between physical/perceptual, geometric, and algebraic realities.

Algebraic invariance way of thinking. An equation can be rewritten in various forms which make certain properties more noticeable. The link between the forms is provided by something which remains invariant, for example the solution set.

Algebraic way of thinking. This is a broad way of thinking. One of its instantiations is the realization that when applying algebra, to a geometry problem for example, one must tell algebra all the relevant geometric constraints. Likewise, we may talk of a geometric way of thinking in this manner; namely, in solving a geometry problem, one must tell geometry all the given conditions.

Usefulness of completing the square.

Classroom Problem 1: What is an ellipse? [See pedagogical discussion of this problem.] Classroom Problem 2: Show that the major axis (the chord containing the foci) of an ellipse is

greater than the minor axis (the chord through the center perpendicular to the major axis).

Classroom Problem 3: Show that any chord AB of an ellipse passing through the center is bisected by the center.

Classroom Problem 4: Find a Cartesian equation of an ellipse. Does it confirm the intuition that an ellipse can be obtained by squashing a circle?

Classroom Problem 5: We know that the tangent line to a circle is perpendicular to the radius at the point of tangency. Is there a similar characterization of the tangent line to an ellipse?

Classroom Problem 6: Is the focus/directrix definition of an ellipse equivalent to the two-focus definition? That is, is the locus of points whose distances from focus and directrix are in the ratio 1e always an ellipse?

Classroom Problem 7: Is a (suitable) plane section of a cone an ellipse? Pedagogical Considerations

We made explicit for our students the overall approach we took in studying circles (global necessity): begin with intuition, formalize a geometric definition capturing it, and represent it

algebraically in a Cartesian coordinate system. We asked students: what other geometric objects might we study in this way? One student suggested an ellipse, for which the intuition is a kind of oval. Another student suggested a formal definition, which the instructor rephrased: Fix two points 1 2,F F in a plane, and a number 1 2d F F . An ellipse is the locus of points P in the plane

Classroom Problem 1: What is an ellipse?

15

such that 1 2PF PF d . The instructor drew one on the board using a shoelace wrapped around two pegs (perceptual verification that the definition is compatible with our intuition).

This problem serves to check that our formal definition can justify properties of the ellipse that we intuitively expect to hold. Let ,T T be the vertices on the major axis with T closer to 2F , A a vertex on the minor axis, and O the center (the midpoint of 1 2F F ). Define

1 2a AF AF so that 2d a . Call b OA and 1 2c OF OF . Then 1 2 2 ,AF AF a also

1 2 2 ,TF TF a and a b . The conclusion would follow from TO a , which would follow from

2 1TF T F . This follows from 1 2 1 2TF TF T F T F upon substituting 2 2TF c for 1TF and 1 2T F c for

2T F .

Another intuition about the ellipse to verify is that any chord AB through the center O should be bisected by the center (symmetry). To prove AO OB , how do we tell geometry what is given? That A and B are on the ellipse is expressed by 1 2 1 2AF AF BF BF . Also, from the definition of the center, 1 2OF OF . Students conjectured that 1 2AF BF but could not justify it; also that 2 1F AF is bisected by AB , which is not true. However, 1 2AOF BOF since they are vertical angles. After several unsuccessful attempts by students, the instructor suggested a proof by contradiction: what disaster follows if, contrary to expectation, AO OB ? Then choose A on OB with OA OA (see Figure 3). From congruent triangles it now follows that both B

and A are on the ellipse; that is, 1 2 1 2A F A F BF BF . This is contrary to intuition, because A is inside 1 2F BF , so the sum of its distances to the vertices 1F and 2F must be less than B 's. The intuition is confirmed by Euclid's Proposition I.21, whose proof the instructor reviewed. The instructor asked students: what is the pedagogical value of this proof? Students suggested: present facts as they are needed in the proof; extract subdiagrams from diagrams and draw them separately; the need for indirect proof was unexpected; the steps must still be assembled into a complete and organized argument [do so!]. The instructor added: sometimes the teacher is an actor, sometimes he is genuinely thinking at the board; the teacher is human and can demonstrate getting stuck and yet recovering. In this case, the instructor knew the status of student conjectures but did not reveal this fact. Part of the important lesson from this problem was that proofs are not done just to have a proof, but also to learn new things along the way.

Classroom Problem 2: Show that the major axis of an ellipse is greater than the minor axis.

c

ba

F2OT' TF1

A

Figure2:Ellipseaxes

Classroom Problem 3: Show that any chord AB of an ellipse passing through the center is bisected by the center.

F2OF1

A

B

A'

Figure3:Bisectedellipsechord

16

Closely related to this problem is the claim that the ellipse is symmetric about its center. We can prove this, and in fact more: the ellipse is symmetric about both axes. Reflection of any point P on the ellipse about the line joining the foci (the major axis) preserves both focal distances and so maps P to another point on the ellipse. Reflection about the perpendicular bisector of the focal segment (minor axis) interchanges the two focal distances and so preserves their sum. The composition of these reflections is reflection about the center, so the ellipse is symmetric about its center.

For this problem, we translate our definition of ellipse into algebra. We will later consider general position and orientation of the ellipse, but for now putting the origin at the center and the foci at ( ,0)c , the definition gives

2 2 2 2( ) ( ) 2 .x c y x c y a (1.1) As 0c this becomes the circle 2 2 2x y a ;

visually, the ellipse seems to be a squashed version of this circle. Indeed, squashing the y coordinate by a factor /b a should produce the correct minor axis, and gives the new equation

2 2 2 2( / ) ( / ) 1x a y b , which a student had remembered. The class therefore conjectured that this equation is equivalent to (1.1). Now there is a Need for Computation to verify the conjecture by reducing (1.1) to this simpler form. The algebra is lengthy but straightforward and uses

2 2 2a b c . A simpler alternative to the brute-force algebra is to introduce notation for the two focal

radii and write equations 2 2 2

12 2 2

2

( ) ,

( ) .

r c x yr c x y

(1.2)

Subtracting gives the simple result 2 21 2 4r r cx , which is productive because we recognize the difference of two squares. Factoring and recalling 1 2 2r r a gives 1 2 2 /r r cx a . Knowing the sum and difference of the focal radii, we can solve for each, obtaining the unexpectedly simple formulas

1

2

,

.

cxr aacxr aa

(1.3)

Substituting back in (1.2) produces the standard form of the equation of the ellipse. This reasoning is simpler and more insightful than the previous computation. Allowing students to see this promotes the Referential Symbolic WoT, which we continued to promote throughout the module. Students should not fear lengthy algebra, but should pause often to unpack the meaning of the symbols and equations, and be alert to the possibility of meaning-based simplification.

The squashed circle viewpoint is very useful for PGA reasoning. For example, we showed that a circle determines its center; does an ellipse determine its foci? The ellipse is obtained by squashing a circle whose diameter is the major axis, in the direction perpendicular to this axis. It follows that the major axis is uniquely determined as the longest chord through the ellipse, and the minor axis as its perpendicular bisector. Their lengths, 2a and 2b , determine c and hence the positions of the foci. (See Supplementary Problem 1 for a Euclidean construction

Classroom Problem 4: Find a Cartesian equation of an ellipse. Does it confirm the intuition that an ellipse can be obtained by squashing a circle?

17

of the foci.)

In our classroom students briefly considered using the radius from the center, but preferred the focal radii. They conjectured that the angles between the tangent line and these may sum to 90 , but then switched to the conjecture that these angles are equal. Students observed that this is true in the special case of a circle, and at the vertices of an ellipse, providing some initial

evidence for the conjecture. Let a line be tangent to an ellipse with equation in standard form at point 1 1( , )P x y in the first quadrant, and make angle with the major axis (see Figure 4). Let the angles between the tangent line and the focal radii iPF be i . Let the angles between

1 2,PF PF and the major axis be , respectively. We want to prove that 1 2 .

Figure4:Characterizationofthetangentlinetoanellipse

Students had several approaches to this problem, each promoting important Ways of Thinking, and the different approaches can be productively compared. First, we need to determine the equation of the tangent line, that is, tell algebra that the line is tangent to the ellipse. This can be done using calculus, or the algebraic method of finding a line having a single intersection of multiplicity two with the ellipse. The result is 2 21 1( / ) ( / ) 1x x a y y b . The notational distinction between 1 1( , )x y , the chosen point of tangency, and ( , )x y , a variable point on the tangent line, deserves emphasis. The Need for Communication requires distinguishing them; using the same symbol for two objects would tell algebra that they are the same. 1 1( , )x y satisfies both the equation of the ellipse and that of the tangent line; ( , )x y satisfies only the latter. Some angles in Figure 4 (more precisely, their tangents) can be computed from the slopes of the lines:

2 21 1

1 1

1 1

tan / ,tan / ( ),tan / ( ).

b x a yy x cy c x

F2OF1

P

Classroom Problem 5: We know that the tangent line to a circle is perpendicular to the radius at the point of tangency. Is there a similar characterization of the tangent line to an ellipse?

18

We can use the exterior angle theorem to find 1 and 2 : 1 and 2 . Then trig identities [prove them!] allow us to compute, for example,

1tan tantan .

1 tan tan

Here is an example of the Necessity for such trig identities (When will I ever use this?). One finds straightforwardly (don't fear complex algebra!)

21 1

21 1

1 21

21

21 1

21 1

2 21

21

tan ,1

( )

tan .1

( )

b x ya y x c

b xa x c

y b xc x a y

b xa c x

We first clear denominators to obtain

2 2 21 1 1

1 2 21 1 1

2 2 21 1 1

2 2 21 1 1

( )tan ,[ ( ) ]

( )tan .[ ( ) ]

b x x c a yy a x c b xa y b x c xy a c x b x

One can check the equality of these by brute force (e.g. multiplying one by the reciprocal of the other and simplifying), but the Referential Symbolic WoT suggests a meaning-based simplification. Indeed, from the equation of the ellipse, the terms 2 2 2 21 1b x a y in the numerators simplify to 2 2a b , and in the denominators 2 2 2a b c can be used. Each tangent then reduces to

2

1

bcy

, so they are equal (note also that referential symbolic reasoning provides students with a

check of their work: since the expression computed is a tangent, it should have no units and thus the numerator and denominator must have the same overall power). Finally, we need to consider whether 1 2tan tan suffices to conclude that 1 2 . This provides Necessity for the concept of a one-to-one function on a specific domain, here 0 .

A second proof of the equal-angles property of the tangent line requires less computation and provides more insight. It suffices to show that the normal line at the point P of the ellipse bisects angle P in the triangle 1 2F PF . If this normal meets the opposite side 1 2F F at a point K , then by Euclid, Prop. VI.3, this is equivalent to showing that 1 2 1 2/ /F K F K F P F P . (An angle bisector in a triangle is characterized by dividing the opposite side in proportion to the adjacent sides.) Since the normal line has equation [Prove this!] 2 2 21 1( / ) ( / )xa x yb y c , the point K has coordinates 2 21( / ,0)c x a and the required proportion becomes

19

2 21 1

2 21 2

( / ) ,( / )

c c x a rc c x a r

which follows from our earlier computation of the focal radii, (1.3). There is a third, physics-based proof which provides insight. Imagine an object moving around the ellipse, with any velocity v , not necessarily constant. Since the sum of its distances to the foci remains constant, the sum of the rates of change of the focal distances is zero. But the rate of change of the distance to a focus is simply (minus) the component of velocity in the direction of the focus. This says that 1 2cos cos 0v v , again implying 1 2 .

Finally, a useful follow-up question is: given an ellipse with its foci, perform a Euclidean construction of the tangent line at a specified point. Besides the solution following from the equal-angles property, namely drawing the focal radii and bisecting the angle they form, there is another following from the squashed circle viewpoint. Begin with the circle 2 2 2x y a and its tangent line at some point 1 1( , )x y . Squashing by a factor /b a in the vertical direction produces the ellipse 2 2( / ) ( / ) 1x a y b along with its tangent line at the corresponding point

1 1( , / )P x by a , and the squashing does not change the x -intercept of the tangent line. Therefore, from the tangent line to the circle we obtain this intercept, and the tangent line to the ellipse is drawn by joining the intercept to P .

The results obtained earlier for the focal radii, e.g.

2 ( / )r a cx a , are surprisingly simple--simple enough that the PGA WoT suggests seeking a geometric interpretation. A linear function of x can be interpreted in terms of the distance from ( , )x y to a vertical line. In this case, 2 ( )r e f x , where we define /e c a and

2 /f a c . That is, 2r is e times the distance from ( , )x y to the vertical line l with equation x f . [We observe that f a , so that this line lies to the right of every point on the ellipse, and also that 1e .] A useful Way of Thinking is that every algebraic computation is a proof of something, and one should try to state explicitly what it proved. Our computation of 2r proved that, if P lies on the ellipse, then 2PF ePl . That is, the distance from P to the focus 2F is e (the

eccentricity) times the distance to l , the directrix. Most students do not yet possess the WoT that the converse should automatically be investigated, and that reversing the algebraic reasoning may prove it. Here we ask whether a point satisfying 2PF ePl is necessarily on the ellipse. If so, then we have an alternative definition of an ellipse, the focus/directrix definition. One way to establish this is to begin with 22 ( )c aa cr x and apply the Pythagorean Theorem (twice) to Figure 5 to obtain

Classroom Problem 6: Is the focus/directrix definition of an ellipse equivalent to the two-focus definition? That is, is the locus of points whose distances from focus and directrix are in the ratio 1e

r2r1

c x F2OF1

P

Figure5:Focus/directrixdefinitionofellipse

20

2 2 2 2 21 2 ( ) ( ) ( )car r c x x c a x . Since 0caa x for every point on the ellipse [prove

it!], we can take the square root to obtain 1 car a x , and add to obtain 1 2 2r r a . Thus, the focus/directrix property implies the two-focus definition of the ellipse.

In 1822, G.P. Dandelin discovered a beautiful proof that conic sections are indeed plane sections of a cone. In our classroom we motivated this

construction and presented the reasoning in the case of the ellipse. It is also possible, but complicated, to take an algebraic approach: obtain equations for a cone and a plane cutting it in three-dimensional Cartesian coordinates, and derive an equation for their intersection. In addition to the complexity of the algebra, a conceptual difficulty is that the three-dimensional Cartesian coordinates ,x y are not also standard Cartesian coordinates in the cutting plane, where we want to have an equation for the locus. Intuition tells us that a horizontal plane cuts a cone in a circle, whereas if the plane is tilted slightly the intersection is an oval curve which may be an ellipse. Perceptually, it seems unlikely that this oval will be symmetric about its center rather than egg-shaped; we need deductive reasoning to correct our intuition in this case. In Figure 6, a cone has vertex V and is

tangent to a sphere along circle C lying in plane and having center O . A plane meets the cone in the curve E (which we assume is closed), is tangent to the sphere at F , and cuts the plane in the line d . Conversely, for any plane cutting the cone in a closed curve E , we can draw a sphere tangent to both the cone and the plane as shown. Our objective is to show that E is an ellipse. P is an arbitrary point on E , the generator VP meets at L , and PM is drawn perpendicular to . PD is drawn perpendicular to d . The cutting angle is PDM and the cone's elevation angle is VLM . It is important for students to attend to the temporal (logical) sequence in which elements are added to the diagram: which points, lines, and so forth are chosen first and which others are then determined by these choices. Some elements are given, such as the cone and plane, some are fixed, like the point of tangency F , and some are variable like

P . We introduce the variable point P in order to show that however it is chosen on E it will satisfy one of the definitions of an ellipse. Students need practice (Repeated Reasoning) with the overall logical structure of such proofs.

Classroom Problem 7: Is a (suitable) plane section of a cone an ellipse?

Figure6:Dandelinsphereconstructionofellipse

21

To prove that E is an ellipse, we need to identify candidates for the focus and directrix. The point of tangency F might be the focus, and d might be the directrix. Some support for this conjecture comes from the facts that, as the plane becomes horizontal, F moves to the center of the circle E and d moves off to infinity. By way of telling geometry that the sphere is inscribed in the cone, we note that PL PF are tangents to the sphere from the same external point. (Students are familiar with the fact that tangents to a circle from an external point are equal, but had difficulty deducing the corresponding fact for spheres. To obtain the latter from the former, one has to find the right circular cross-section of the sphere.) We also have

/ / sinPM PL PM PF and / sinPM PD . Therefore, sin ,sin

PF ePD

a constant independent of P . If we know that 1e , then we have shown that E is an ellipse. This follows from our implicit assumption that the plane cuts the cone in a closed curve, which occurs only when the plane cuts each generator of the cone. That in turn requires , which implies 1e .

Although Dandelin's proof is visual and elegant, it also raises several subtle issues which should be made explicit if not necessarily resolved rigorously in class. First, how do we know that there is always a (unique) sphere tangent to both the cone and the plane ? An intuitive continuity argument is convincing but not rigorous: begin with a small spherical balloon tangent to the cone only, and ``blow it up" until it becomes tangent to the plane as well. A geometric proof can be given by using a suitable plane cross-section of the three-dimensional diagram and the fact that a triangle has a unique inscribed circle. Second, the fact that a cutting plane with produces a closed curve of intersection relies on some intuitive topology. (Indeed, the proof shows only that the curve of intersection is contained in an ellipse. We are relying on this intuitive topology for the conclusion that the curve is a complete ellipse.) We did not press the issue of what it means to be a closed curve at this point, but we returned to it when comparing the Dandelin sphere proofs for the different conic sections. Third, the definition of a cone as a figure of solid geometry may still be at the intuitive level for students and should be formalized. For example, begin with a circle and a point other than its center on the line through its center and perpendicular to its plane; take the union of all lines joining this point to the points of the circle. Students do not always realize initially that this defines a double cone.

Supplementary and Practice Problems: Problem 1. Given an ellipse, construct (synthetically) its foci. Although the construction is purely synthetic, the proof uses coordinates. First, we find the center. Construct a pair of parallel chords (AB and CD in Figure 6). Connect the midpoints

1M and 2M . Construct a second pair of parallel chords EF and GH not parallel to the first pair, and connect their midpoints 3M and 4M . Then the intersection of 1 2M M and 3 4M M is the center of the ellipse. To prove this, it suffices to show that the center is contained on the line joining midpoints of any 2 parallel chords. Let the ellipse be given by 2 2( / ) ( / ) 1x a y b . Suppose one chord is given by the line 1y mx n and the parallel chord is given by 2y mx n . Then the endpoints of the first chord are found by solving 2 2 2 2 2 2 2 21 12 ( ) 0b m a x mn a x n a b . However, we can find the midpoint without

22

finding the endpoint: its x-coordinate will be 2

11 2 2 2

mn axb m a (average of 2 solutions to a

quadratic is / 2 )B A , so y-coordinate is 2 2

11 12 2 2

m n ay nb m a . Similarly, the other midpoint will

be 2 2 2

2 22 2 22 2 2 2 2 2( , ) ,

mn a m n ax y nb m a b m a

. To show that the line connecting the midpoints

contains the origin, we verify that 1 21 2

y yx x .

2 2 2 2 2

2

2 2 2 2 2

2

2

2

( )

( )

,

i i i

i i

y m n a n b m ax mn a

m a b m ama

bma

which is independent of in , giving the desired equality. To find the axes, we pick any point on the ellipse and draw a circle centered at O through that point. The circle will intersect the ellipse at 4 points, which we connect to get a rectangle whose sides are parallel to the desired axes [prove this!]. We can then draw lines parallel to these sides through the origin. With the axes in hand, pick a point P on the

ellipse that is also on the minor axis. Draw a circle centered at P with radius equal to the semi-major axis. The intersection points of this circle with the major axis are the foci. The most difficult step in this proof can be viewed as showing that the locus of midpoints of a family of (all possible) parallel chords of an ellipse is a diameter of the ellipse. Analogous statements for the parabola and hyperbola can be proved similarly. Problem 2. Use a second Dandelin sphere to confirm that an ellipse has a second focus/directrix pair. Problem 3. Let E be an ellipse with center O , major axis of length 2a , and minor axis of length 2b . Prove that if P E , then b PO a . Problem 4. Find the equation of the ellipse with foci 2 23 3,0 and ,0 and directrix the line

4x . Problem 5. Find the eccentricity and the directrices of the ellipse

2 22 3 1.5 4

x y Problem 6. Consider the equations

M4

M3

M2

M1

O

A

B

C

D

E

F

G

H

Figure7:Constructingellipsefoci1

23

2 2

2 2

4 5 16 10 39 04 5 16 10 39 0.

x y x yx y x y

a. Show that one of the above equations defines an ellipse and that the other equation does not define an ellipse. b. For the equation which does define an ellipse, find the coordinates of the center, the coordinates of the foci, the length of the major axis, and the length of the minor axis.

Problem 7. Find an equation for an ellipse with the same foci as the ellipse 22

25 16 1yx but

greater eccentricity. Graph both ellipses on the same set of axes. Problem 8. Let E be an ellipse with center O . Let P E . Find an expression for PO in terms of the angle between PO and the major axis of E . Problem 9. Find a tangent line to the ellipse 2 24 196x y which is perpendicular to the line

5x . Problem 10. Find a normal line to the ellipse 2 23 2 50x y which is perpendicular to the line 6 7x y . Problem 11. Prove that if PQ is a diameter of an ellipse, then the tangent lines to the ellipse at the points P and Q are parallel. Problem 12. Let PQ be a diameter of an ellipse.

a. Prove that the midpoints of the chords of the ellipse which are parallel to PQ all lie on the same line.

b. Prove that the line containing the midpoints of the chords of the ellipse which are parallel to PQ is parallel to the tangent lines to the ellipse at the points P and Q .

Problem 13. Is a plane section of a circular cylinder also an ellipse? Perceptually, it seems unlikely that sections of cylinders and cones have the same shape. The Dandelin sphere argument actually applies to the cylinder as well as the cone. Again, this is an excellent opportunity for students to appreciate the advantages of the Deductive over the Empirical proof scheme. Problem 14. A mining company drills a mineshaft into the ground. The hole is a circular cylinder, with its axis making an angle with the vertical. If the (elliptical) opening at the surface must fit within the square plot of ground owned by the company, whose side is twice the diameter of the cylinder, how does this restrict ? If the axes of the ellipse are assumed parallel to the sides of the square, this is an easy application of the squashing of the circle to an ellipse. However, a larger ellipse will fit diagonally. The sides of the square are then tangent to the ellipse and make a 45 degree angle with the axes. Determining where such lines are tangent is a nontrivial exercise. If we tilt our heads by the angle , we see the cylindrical mineshaft as if its axis were vertical, and the ground as a cutting plane with elevation angle . Then the elliptical opening can be described by the equation 2 2 2 2cosx y r , where r is the radius of the hole. Here the ellipse is obtained by stretching, rather than squashing, a circle. In the simplest approach to the problem, one might assume that the ellipse's axes are parallel to the sides of the square, so the constraint is simply that the major axis 2 / cosr not exceed the side 4r of the square. This gives cos 1 / 2 , or / 3 . However, a larger ellipse will fit in the square with the major axis along the diagonal. Rather

24

than inscribe a tilted ellipse inside the square, it may be easier to draw a tilted square around the standard ellipse. The sides of the square, with slopes 1 , must then be tangent to the ellipse. To achieve this we use implicit differentiation to locate the points where the ellipse has slope 1 , finding 22 cos 2 0x yy with 1y , so that 2cosy x . (Of course, other methods of finding the tangent lines, discussed above, lead to the same result.) Substituting back into the equation of the ellipse gives the points of tangency as

2

2( , ) ( 1, cos ),

cos 1 cosrx y

where all four sign combinations are possible. The x -intercept of the tangent line at the point with positive coordinates can be computed as

21 cos .cos

r The four tangent lines form a square whose diagonal is twice this value, so its side is this value times 2 . The problem constraint is then

22 1 cos 4 .cosr r

which can be solved to yield simply cos 1 / 7 0.37796. This translates into o67.79 , versus 060 as found previously when the major axis was parallel to the sides. Problem 15. A dog is tied up in a large field by a loop of rope running through its collar and around two posts. If the posts are separated by a distance d , and the rope has length l , this restricts the dog to an elliptical region in which the sum of focal distances is less than l d . If a straight river runs through the field along a given line, how close to this river can the dog get? Consider a numerical example in which 20l and 6d . Let the ellipse be centered at the origin with major axis along the x -axis, while the river has equation 16y x . Then

7, 3,a c and the equation of the ellipse is 2 2( / 49) ( / 40) 1x y . As in Problem ellipse9, there are two tangent lines to the ellipse parallel to the river, namely 89y x . The minus sign gives the tangent line nearer to the river, and the distance between this line and the river is the closest the dog can get. The distance between parallel lines of slope m whose y -intercepts

differ by b is 2/ 1b m , which in our case gives a distance of (16 89) / 2 4.6429 .

25

Unit 3: The Parabola Focus WoT's and WoU's Understanding the way the equation of a locus changes depending on how it is placed

relative to the coordinate system, and that algebraic transformations induce a translation of the locus.

Structural way of thinking. Try to generalize established properties and explore a parameter space (if an ellipse has 0 1e , what happens for 1e ?).

PGA way of thinking: Attending to interrelationships between physical/perceptual, geometric, and algebraic realities.

Algebraic invariance way of thinking. An equation can be rewritten in various forms which make certain properties more noticeable. The link between the forms is provided by something which remains invariant, for example the solution set.

Usefulness of completing the square. Classroom Problem 1 What locus satisfies the focus/directrix definition with 1e ? Classroom Problem 2 How does the equation obtained for the parabola depend on the placement of the focus and directrix relative to the chosen coordinate system? Which placement leads to the simplest equation? Algebraically, how can the other equations be transformed into this one? Pedagogical Considerations:

Translating the definition PF Pl into algebra yields various equations depending on where the focus and directrix are placed relative to the chosen coordinate system. Examples that were discussed in our class include:

Focus ( ,0)k , directrix y -axis: 2 22y kx k . Focus ( ,0)k , directrix x k : 2 4y kx . Focus (0, )k , directrix x -axis: 2 22x ky k , or 2 2( ) / 2y x k k . Focus ( / 2,0)k , directrix / 2x k : 2 2y kx . Focus (0,0) , directrix 0Ax By C , assuming without loss of generality that 2 2 1A B

: 2 2 2 2 22 2 2 .B x A y ACx BCy ABxy C

Students should see a need to reconcile these equations with their previous understanding of a parabola, probably defined as the graph of a quadratic function 2y ax bx c . Only one of the above equations has this form, but students should recognize that reversing the roles of x and y all but the last to be put into this standard form. The last example, with a general line as directrix, creates Necessity for the formula for the distance from a point (focus) to a line (see line unit).

One pair of students in our classroom interpreted this problem geometrically rather than algebraically, and they presented a geometric construction of points on the locus. Given the focus F and directrix l , choose a point Q on l , draw FQ and mark its midpoint M . Then the

Classroom Problem 1: What locus satisfies the focus/directrix definition with 1e ?

26

perpendicular bisector of FQ and the perpendicular to l at Q meet at a point P which is on the locus (see Figure 8). The required condition FP QP follows from

FMP QMP , by SAS. One can use this construction to see that the locus is symmetric about an axis through F and normal to l . Another student translated this construction directly into algebra. Taking F to be (0, )k and l to be the x -axis, let Q be at 0( ,0)x . One easily calculates that M is

0( / 2, / 2)x k and P is at ( , )x y where

0x x and 2 20( ) / 2y x k k . Eliminating 0x yields 2 2( ) / 2y x k k , one of the previously calculated equations for a parablola.

This connection between geometry and algebra stimulated a discussion of whether the constructive approach has any pedagogical advantage compared to the direct derivation of the algebraic equation from PF Pl , which seems easier. A student suggests that it requires, and shows, greater understanding of the construction and the definition of parabola. It promotes the PGA WoT by showing that the connection between geometry and algebra can be made at a variety of stages of a computation or levels of analysis. It also provides Necessity for confronting the distinction between a Cartesian equation and a parametric representation of a locus. The construction naturally gives a parametric representation, parametrized by 0x , the coordinate of the arbitrary point Q . Coincidentally, this problem is so simple that the parametric representation immediately reduces to the Cartesian version.

The intent of this problem is for students to develop the technique of translating axes to simplify the equation of a curve. (They can also interchange x and y to reflect about the line y x , but we postpone the use of rotations of axes until later.) Students must carefully reason through the process to avoid subtle sign errors. First, it is necessary to

decide whether the translations will be active or passive: are we moving the locus on a fixed coordinate system, or are we moving the coordinate axes while keeping the same locus? In our classroom we chose the former. How do we modify the equation ( , ) 0f x y in order, say, to move the curve a distance a in the x direction? One should avoid the notation x x a unless its meaning has been thoroughly discussed and agreed upon. Does it mean that each point

( , )P x y on the curve is moved to the new location ( , )P x a y , or does it mean that we modify the equation by substituting x a for each occurrence of x ? In fact, if we move each point as stated, the new curve has equation ( , ) 0f x a y . Repeated reasoning will be needed for students to understand the source of the sign change. One possibility is to introduce

,x x a y y for the coordinates of the moved point, so that ( , ) ( , ).f x y f x a y

l

P

M

F

Q

Figure8:Geometricconstructionofpointsonparabola

Classroom Problem 2: How does the equation obtained for the parabola depend on the placement of the focus and directrix relative to the chosen coordinate system? Which placement leads to the simplest equation? Algebraically, how can the other equations be transformed into this one?

27

Supplementary and Practice Problems: Problem 1:

c. Let C be the set of points in the plane satisfying the equation ( , ) 0f x y . Let C be the set of points obtained by translating the points in C up 3 units and to the right 5 units. What equation is satisfied by the points in C ?

d. Let C be the set of points in the plane satisfying the equation ( , ) 0f x y . Let C be the set of points obtained by translating the points in C down 3 units and to the right 5 units. What equation is satisfied by the points in C ?

e. Let C be the set of points in the plane satisfying the equation ( , ) 0f x y . Let C be the set of points obtained by translating the points in C up 3 units and to the left 5 units. What equation is satisfied by the points in C ?

f. Find an equation defining the parabola obtained by translating the parabola 2 2 0y kx down m units and to the left n units.

Problem 2: Find the focus and the directrix of the parabola 22y x . Problem 3: Find the focus and the directrix of the parabola 22 16 31y x x . Problem 4: Give an example of an ellipse and a parabola with the same focus and directrix. Problem 5: Find a tangent line to the curve 25 0y x that is perpendicular to the line 20 1x y . Problem 6: Prove that an appropriate intersection of a plane and cone gives a parabola. How does the Dandelin sphere construction for the parabola compare to that for the ellipse? In addition to considering how to place the Dandelin sphere, students should also realize that the double Dandelin sphere argument does not apply. Problem 7: Show that the tangent line to a parabola at a point P makes equal angles with the line joining P to the focus and the line through P perpendicular to the directrix. Problem 8: (Putnam Exam) A dart is thrown at a square dartboard, with equal probability of landing anywhere on it. What is the probability that it lands closer to the center than to any edge? Take the square to be bounded by the lines 1x and 1y , so that the center is at the origin and the area is 4 . The locus of points equidistant from the center and the side 1y is the parabola having these as focus and directrix; its equation is 2(1 ) / 2y x . Rotations by 90 degrees give the parabolas corresponding to the other sides. The dart lands closer to the center than to any edge if it lands in the closed region bounded by these parabolas, roughly an octagon with curved sides. Using the octagonal symmetry, the area is 8 times that of the segment in the first quadrant bounded by the y -axis, the parabola 2(1 ) / 2y x , and the line y x . The probability is obtained by dividing by 4 , the area of the square, and is

2 1 2

0(1 2 ) (4 2 5) / 3 0.219.x x dx

Unit 4: The Hyperbola Focus WoT's and WoU's Determining tangent lines and asymptotes. Referential symbolic reasoning: simplifying the equation of a locus by translation of axes. PGA WoT.

28

Structural way of thinking. Try to generalize established properties and explore a parameter space (if an ellipse has 0 1e , what happens for 1e ?).

Usefulness of completing the square. Classroom Problems Classroom Problem 1: What locus satisfies the focus/directrix definition with 1e ? Classroom Problem 2: What are the asymptotes of a hyperbola? Does a parabola have asymptotes? Classroom Problem 3: (a) Find a condition on , , ,a b m n so that the line y mx n is tangent to the hyperbola 2 2 2 2( / ) ( / ) 1x a y b . (b) Prove that the tangent line to a hyperbola bisects the angle between the focal radii. (c) Find the angle between a hyperbola and an ellipse that share the same foci. Pedagogical Considerations

By this point, students should pose this question themselves. Based on their experience, our students conjectured that the locus is a hyperbola, meaning a curve with equation 2 2 2 2( / ) ( / ) 1x a y b . To verify

this, students suggested a coordinate system in which the focus F is at ( ,0)k and the directrix l is along the y -axis. Then the defining property PF e Pl immediately gives

2 2 2 2( ) .x k y e x We now wish to simplify this equation to a symmetric form like that of the ellipse, for example making any symmetry of the locus obvious. The following computation is remarkable in that no high school student would recognize it as simplification. The equation gets more complicated at each step, and even the final form is about as long as the initial form. It counts as simplification only because we possess the PGA Way of Thinking that translating the coordinate axes and introducing new parameters for combinations of e and k will clarify the geometry of the locus. Expanding, collecting terms, and completing the square in x leads to

2 22 2 2

2 2( 1)( ) .1 1k k ee x y

e e

Introducing 2 2 2 2 2/ ( 1)a k e e and 2 2 2 2/ ( 1)b k e e (since both quantities are positive) leads to

22 2

12 2

( )1.

ke

x ya b

This tells us that there is a better choice of coordinates than the one we made: we should transform 2 1

ke

x x and y y . The arrows mean that each point ( , )x y on the curve should be moved right to the location given by the shifted coordinates. The final form of the equation of the hyperbola is indeed 2 2 2 2( / ) ( / ) 1x a y b . Its x -intercepts are ( ,0)a .

What can be said about the precise shape of the hyperbola whose equation we have found? It is now clearly symmetric about both the x and y axes (hence the origin too) and

therefore its shape is determined by the portion in the first quadrant, 2 2( / )y b a x a . Perhaps it is monotonically increasing, like x ? Can we prove this? What does it mean? A

Classroom Problem 1: What locus satisfies the focus/directrix definition with 1e ?

29

student suggested that larger x 's give larger y 's. How can we tell algebra this meaning? One

possibility is that if 1 2x x ,then 1 2( ) ( )y x y x ; that is, 2 2 2 21 2( / ) ( / ) .b a x a b a x a This is true because the composition of the increasing functions 2 2x a and x is increasing. Using the derivative to test for an increasing function would be overkill; calculus is not needed here. We need to locate the focus and directrix of the hyperbola given by our final equation, which have been lost in translation (bad pun). Since the shift right was by /a e , the directrix is now the line /x a e . Applying the same shift to the focus locates it at ( ,0)ae . Note that /a e a ae . Because of the symmetry of the curve, there is a second focus at ( ,0)ae and a corresponding directrix at /x a e .

In our classroom we distributed a textbook derivation of the equation of the hyperbola to our students for comparison with their solution to this problem. The textbook derivation employs 20/20 hindsight to situate the focus at ( ,0)ae and the directrix at /x a e from the beginning, miraculously resulting in the simple standard form of the equation. We contrasted this with our implementation of the Necessity Principle and the PGA WoT. Students should see for themselves the effect of various placements of the focus and directrix against the coordinate axes and decide which they consider simplest.

One neat approach to discovering the asymptotes of the hyperbola is to look at the limiting behavior of its tangent lines as the point of tangency moves off to infinity (see Figure 9). The line tangent to the hyperbola at 1 1( , )x y has the equation

1 1

2 2 1.xx yya b

(1.4)

This can be rewritten 2 2

12

1 1

,b x ay xa y y

showing the slope and x-intercept 2

1

,0aKx

. The x-

intercept clearly approaches the origin as our point

moves to infinity; the form of the slope necessitates

examining the ratio 11

xy

. From the equation of the

hyperbola, one obtains 21

11

1 ( / ) .x a b yy b

Thus, as 1y , we must have 1x also, in such a way that 1 1/x y has the limiting value /a b . Therefore, the tangent line (1.4) has the limiting form ( / )y b a x , and the asymptotes are these lines through the origin (center). One can show also that the vertical (or

Classroom Problem 2: What are the asymptotes of a hyperbola? Does a parabola have asymptotes?

4

2

-2

-5 5

K

P

F2F1

Figure9:Limitingbehavioroftangentlinetoahyperbola

30

perpendicular) distance from the hyperbola to an asymptote goes to zero in the limit. To see the three-dimensional origin of the asymptotes, take a plane cutting a cone in the

hyperbola, and a parallel plane through the vertex cutting it in a pair of lines. The hyperbola meets every generator of the cone except for these two lines. They should be related to the asymptotes, but unfortunately do not lie in the plane of the hyperbola. Projecting them into this plane, and translating them to pass through the center, indeed gives the asymptotes. A parallel treatment of the parabola shows that it has no asymptotes. (It also has no center.) The tangent line to the parabola 2 2y kx at 1 1( , )x y has the equation 1 1( )yy k x x , which can also be written

1

1

.2

k yy xy

In the limit 1y , the slope goes to zero and the intercept to infinity; the tangent line has no limiting position.

This type of problem should be straightforward by now. The tangency condition in part (a) is 2 2 2 2a m n b . The equal-angle property of the hyperbola can be established in any of the ways used for the ellipse. For part (c), if an ellipse and a hyperbola with the same foci intersect at P , then the

tangent line to the hyperbola at P coincides with the normal to the ellipse there, because both bisect the same angle between the focal radii (See Figure 10). Therefore the two curves are orthogonal at P . Students may not realize without reflection that neither the ellipse nor the hyperbola is completely determined by specifying their foci; one more parameter such as the eccentricity is required. Thus we have a whole family of ellipses and hyperbolas, and all are orthogonal when they intersect!

Supplementary and Practice Problems: Problem 1: Find the equation of the hyperbola with foci ( 5,0) and (5,0) and directrices the lines

8

6

4

2

-2

-4

-6

-5 5

K

P

F2F1

Classroom Problem 3: (a) Find a condition on , , ,a b m n so that the line y mx n is tangent to

the hyperbola 2 2 2 2( / ) ( / ) 1x a y b . (b) Prove that the tangent line to a hyperbola bisects the angle between the focal radii. (c) Find the angle between a hyperbola and an ellipse that share the same foci.

Figure10:Ellipseandhyperbolawiththesamefoci

31

16 / 5x . Problem 2: Find the center, eccentricity, foci, directrices, and asymptotes of the hyperbola

2 22 4 3 0.x x y Problem 3: Let 1F and 2F be the foci of the hyperbola

22

2 2 1yxa b . Show that P is a point on the hyperbola if and only if 1 2| | 2 .PF PF a Problem 4: Prove that an appropriate intersection of a plane and cone gives a hyperbola. How does the Dandelin sphere construction for the hyperbola compare to that for the ellipse? Problem 5: Let the tangent line to a hyperbola at a point P meet the asymptotes at points ,A B . (a) Show that P is the midpoint of the segment AB . (b) Show that the area of the triangle ABO , where O is the center of the hyperbola, is independent of which point P was chosen. Problem 6: The Greeks (Pappus) used conic sections to trisect angles. Reconstruct their method, as follows. (a) Given a line segment AB , show that the locus of all points P such that 2PBA PAB is

(a branch of) a hyperbola. Identify its focus, directrix, and eccentricity. (b) An angle AOB can be trisected as follows. Assuming without loss of generality that

OA OB , draw a circle with center O passing through A and B . Next draw the hyperbola determined as above by segment AB . Let it cut the circle at a point P in the interior of the angle. Show that 3 POB AOB .

32

Unit 5: The Line

Focus Ways of Thinking and Ways of Understanding PGA way of thinking Algebraic invariance way of thinking

The line is an object that is so familiar to students, both perceptually and mathematically, that they are likely to take the algebraic characterization of a line for granted (Of course the equation y mx b defines a line!). For this reason, we do not place a unit on the line at the beginning of

this modulestudents would not understand the questions being asked at that point. However, the preceding processes of characterizing various conic sections should have prepared these students to recognize that the following problems are indeed legitimate questions. Classroom Problem 1: Find the equation of a line. Classroom Problem 2: Show that if A and B are both non-zero, then the equation

0Ax By C defines a line. Pedagogical Considerations

Here we list three approaches to the problem. The first approach uses similar triangles. Consider a line l . In order to tell algebra about the line l , we should choose a coordinate system; we will choose this system in such a way that l is not

parallel to either of the coordinate axes. A line is determined by two points; call them 1 1( , )P x y and 2 2( , )Q x y . Let ( , )R x y be any other point on the line l . Without loss of generality, we may assume that 1 2x x x . Let S and T be the points with coordinates 2 1( , )x y and 2( , )x y , respectively. Since RT

is parallel to the y axis and QT is parallel to the x axis, QTR is a

right angle. Similarly, PSQ is a right angle. Moreover, QRT and PQS are corresponding angles formed by the parallel lines RT

and QS

and the transversal l and are therefore

congruent. Thus QRT is similar to PQS by AA similarity. It follows that QS RTPS QT , that is: 2 1 2

2 1 2

.y y y yx x x x

Naming this common ratio m and solving for y , we find that y mx b , where 2 2b y mx . Students should be able to explain why generalizing this argument implies that any point ( , )x y on the line satisfies the equation y mx b . They should also realize the reason for fixing two points that define the line, which we denote by particular coordinates, and considering any other point on the line, which we denote with general coordinates ( , )x y .

A second approach is to suppose that the given line l is the perpendicular bisector of some segment PQ . If the coordinates of P and Q are 1 1( , )x y and 2 2( , )x y , respectively, and ( , )x y is a point on l , then since ( , )x y is equidistant from P and Q , we have

2 2 2 21 1 2 2( ) ( ) ( ) ( )x x y y x x y y , or equivalently, 0Ax By C , where

2 12 2A x x , 2 12 2B y y , and 2 2 2 21 2 1 2C x x y y . A third approach (requiring some familiarity with the notion of betweenness; P~R~Q

Classroom Problem 1: Find the equation of a line.

33

means R is between P and Q) is to take the concept of distance as primary and define a line as having distances sum. That is, given points P and Q, we define PQ

as the union of 3 sets:

1. { : ~ ~ & }R R Q P RQ QP RP 2. { : ~ ~ & }R Q R P RQ RP QP 3. { : ~ ~ & }R Q P R QP PR QR The necessity for all 3 of these sets can be established through overaching problem 8. If one blindly applies the distance formula, the calculation is lengthy. However, we can choose (say) the x-axis to be along the direction of PQ

, in which case it is clear that R has the desired

properties iff it has y-coordinate of 0. From an expert perspective, this is really defining a line through defining an axis. That is, we define a primary line as the set of all ordered pairs (x,0), or y=0. From this, we can apply translation and rotation to obtain the general for