1 Supplementary Modules for Pre-Service Mathematics Teachers Guershon Harel, Jeff Rabin, Laura Stevens, Evan Fuller University of California, San Diego 2 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 communication—in this case the need to communicate to others a precise description of the curve formalizing the intuition—demands 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 certainty—to 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 true—the cause that makes it true—a 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 structure—the need for structure. Specifically, our aim is to lead students to recognize a common structure among all the conic sections—geometric 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 people’s current knowledge through engagement in problematic situations conceived as such by them. Psychological need, on the other hand, has to do with people’s desire, volition, interest, self determination, and the like. Indeed, before one immerses 3 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 internalization—a 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 thinking—a 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 matter—definitions, 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 lim f ( a h ) f ( a ) / h ” or “the instantaneous rate of change h0 4 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. you learned that a circle with center ( a . The link between the forms is provided by something which remains invariant. b ) and radius r is represented by the equation ( x a )2 ( y b)2 r 2 . Usefulness of completing the square. Tami expanded this equation into the equation x 2 y 2 2 x 4 y 2 0. to a geometry problem for example.1 Focus Ways of Thinking and Ways of Understanding PGA way of thinking: Attending to interrelationships between physical/perceptual. We draw a circle and ask.5 1. Write down your meaning of the statement “a circle with center ( a . Now consider the circle ( x 1)2 ( y 2)2 3 . An equation can be rewritten in various forms which make certain properties more noticeable. we can rewrite the above equation in the form x 2 y 2 2ax 2by ( a 2 b2 r 2 ) 0 . b ) and radius r . or at any other point. ( x a )2 ( y b)2 r 2 or as . 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. and algebraic realities. Likewise. This could be communicated by specifying the center and radius (and the plane. Can Bruce retrieve the center and radius of the circle from the second equation? Classroom Problem 3: In the previous two problems. b ) and radius r has the equation ( x a )2 ( y b)2 r 2 . in solving a geometry problem. We begin with an object familiar to students both perceptually and mathematically: the circle. Classroom Problem 1: You drew a circle with a compass but forgot to mark its center. Algebraic invariance way of thinking. geometric. Understanding a property derived from this definition: the tangent line at a point is perpendicular to the radius to that point. The following problems may be used in class to solidify these ideas through the necessity principle and the repeated reasoning principle. Algebraic way of thinking. This is a broad way of thinking. This shows that the equation of a circle can be written in two different forms: as 1. if not obvious).1 The Circle 1.1.” Explain again why this is the case. 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 ( x a )2 ( y b)2 r 2 represents a circle with center ( a . namely. we may talk of a geometric way of thinking in this manner. The problem that Bruce encountered can be generalized: If we expand the brackets in the circle’s equation and collect the corresponding terms. one must “tell geometry” all the given conditions. This definition translates easily into the usual Cartesian equation for a circle centered at the origin. Bruce didn’t see the original equation. One of its instantiations is the realization that when applying algebra. Understanding the concept of tangent line as a line intersecting a circle only once. for example the solution set. one must “tell algebra” all the relevant geometric constraints. 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. P3 on the circle. We can make the second equation more compact and easier to remember by substituting f 2a . 2) . Each of these bisectors therefore passes through the center. indeed. However. How would you define “angle between two intersecting planes”? 2. The two solutions are mathematically the same. The locus of points equidistant from any two of the points is the perpendicular bisector of the line segment joining them. Classroom Problem 6: A circle of radius 4 is centered at (1. This solution is obviously related to the theorem that the perpendicular bisectors of the sides of any triangle are concurrent. We often talk about orthogonal lines.2 Pedagogical Considerations A recurring theme of this unit (and.1. which can be found as the intersection of any two of them.” 1. Question: Does every equation of the form x 2 y 2 fx gy h 0 represent a circle? Classroom Problem 4: You are familiar with the concept “angle between two intersecting lines. x 2 y 2 fx gy h 0. all the units of this module) is that both geometry and algebra are systems for drawing logical conclusions from given data. Where does it cross the coordinate axes? What lines are tangent to it at these points? Classroom Problem 7: Prove that the intersecting circles x 2 y 2 f1 x g1 y h1 0 and x 2 y 2 f 2 x g 2 y h2 0 are orthogonal if and only if f1 f 2 g1 g 2 2(h1 h2 ). we use the locus definition of the circle: the three points are equidistant from the center. x 2 y 2 2ax 2by ( a 2 b2 r 2 ) 0 . 1. and all must be used nontrivially in the reasoning. Can we talk about orthogonal curves? Classroom Problem 5: Construct two orthogonal circles in one plane. This shows that every circle can be represented by an equation of the form 3.6 2. and a line orthogonal to a plane. To exploit the power of these systems by drawing the strongest conclusions. 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. of course. A geometric solution to this problem begins by choosing three points P1 . This is what we referred to earlier as the algebraic way of thinking and geometric way of thinking. Because three (noncollinear) points determine a circle. It is possible to retrieve the center of the circle? . in fact meeting at the center of the circumscribed circle. these three points suffice to “tell geometry” which circle is given. How would you define “angle between two intersecting curves”? 4. and h a 2 b 2 r 2 to obtain x 2 y 2 fx gy h 0. How would you define “angle between a line and plane” (if they intersect)? 3. g 2b . 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. rather than some other curve. P2 . orthogonal planes. To tell geometry that we want the circle through these three points. Traditionally. Bruce didn’t see the original equation. This is the algebraic proof of the claim that the perpendicular bisector is the locus of equidistant points. the specific distance r having cancelled out (referential y y2 x x x x 1 2 a 1 2 . for example. Can Bruce retrieve the center and radius of the circle from the second equation? 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. b ) lies on the perpendicular bisector of the segment PP 1 2 . Our experience suggests that this meaning is not always clear to students. Students may wish to locate the center as the intersection of two diameters. Tami expanded this equation into the equation x 2 y 2 2 x 4 y 2 0. yi ) . Given a ruler. 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. The conditions telling algebra that the three points lie on a circle with center ( a . There is an algebraic solution mirroring this geometric one. for example. In our approach to this module. Having an equation for some locus is so powerful that students are unlikely to question the correspondence between the locus and the equation—they see deriving an . Subtracting a pair of these equations gives a condition that the center is equidistant from two of the given points. Write down your meaning of the statement “a circle with center ( a . b ) and radius r are the three equations ( xi a ) 2 ( yi b) 2 r 2 . Now consider the circle ( x 1)2 ( y 2)2 3 . The center can be found by solving any two of these linear equations simultaneously. and therefore the center lies on it too. thus making the fold line a diameter. This turns out to be. b ) and radius r . 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. It is useful for students to realize how geometric solutions depend on the available tools.7 construction—that there was a reason to think of the two chords. it can be done by inscribing a right angle at a point of the circle. b ) and radius r is represented by the equation ( x a )2 ( y b)2 r 2 . (Is this an allowed use of a ruler?) With straightedge and compass.” Explain again why this is the case. How easy this is depends on the available tools. the ordered pair of every point on the curve satisfies the equation. The perceptual approach is valuable for highlighting intuitive properties of the circle. and conversely. its sides will meet the circle at ends of a diameter. b 1 2 2 y2 y1 precisely the statement that ( a . Let the point Pi have coordinates ( xi . This is an excellent example of an empirical/perceptual proof as opposed to a deductive one. symbolic reasoning). sometimes assuming that a chain of reasoning proves an “if and only if” claim even though the reasoning is not reversible. This issue goes beyond the phenomenon that students are not careful to distinguish a statement from its converse. we aim at demystifying mathematics for the students. Two such diameters intersect at the center. Classroom Problem 2: Some of the solutions to the previous problem led us to conclude that an equation of the form ( x a )2 ( y b)2 r 2 represents a circle with center ( a . The third linear equation is dependent on the two chosen. They state that each point is r units from the center. one might locate two points on the circle a maximal distance apart and join them. This could necessitate the theorem that an inscribed angle measures half of the intercepted arc. which should be deductively verifiable later: its symmetry about any diameter. However. This productive disequilibrium is a very important experience in terms of the origin of mathematical definitions. new concepts often rely on old ones: the angle between planes is defined in terms of the already-understood angle between lines.” 1. or else empty. First. and a line orthogonal to a plane. x 2 y 2 fx gy h 0. Question: Does every equation of the form x 2 y 2 fx gy h 0 represent a circle? equation as the end. we gave it repeated attention when working with the different conic sections. We can make the second equation more compact and easier to remember by substituting f 2a . Again students completed the square to recast the equation as f g f 2 g2 x y h . you learned that a circle with center ( a . To help students understand this issue. 2 2 For the angle between two planes. This shows that the equation of a circle can be written in two different forms: as 1. and h a 2 b 2 r 2 to obtain x 2 y 2 fx gy h 0. This 2 2 4 represents a circle iff the constant on the right side is a positive number r 2 . They realized that this concept is not well-defined. 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. that is iff f 2 g 2 4h . students suggested choosing a point in the intersection of the planes. This shows that every circle can be represented by an equation of the form 3. orthogonal planes. we can rewrite the above equation in the form x 2 y 2 2ax 2by (a 2 b2 r 2 ) 0 . How would you define “angle between a line and plane” (if they intersect)? 3. if a definition allows different people to make different choices. Can we talk about orthogonal curves? . The problem that Bruce encountered can be generalized: If we expand the brackets in the circle’s equation and collect the corresponding terms. and taking the angle between these lines. How would you define “angle between two intersecting curves”? We often talk about orthogonal lines. This promotes the Algebraic Invariance way of thinking and introduces the usefulness of completing the square. g 2b . an essential technique throughout this module (Repeated Reasoning). choosing a line through this point in each of the planes. They decided instead to use the angle between the planes’ normal lines. x 2 y 2 2ax 2by (a 2 b2 r 2 ) 0 . which is not ambiguous.8 Classroom Problem 3: In the previous two problems. which lead to different results. Students should ask (or be asked) what is the locus if this condition is not met: a single point. How would you define “angle between two intersecting planes”? 2. then the Need for Communication requires that these choices be standardized in some way so that everyone obtains the same answer. 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. b ) and radius r has the equation ( x a ) 2 ( y b) 2 r 2 . because it depends on the choices made. ( x a )2 ( y b)2 r 2 or as 2. Similarly. 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. and the case of a single intersection defines the idea of a tangent line. or no intersection points. one. this would advance another important way of thinking—that a term can have multiple interpretations. students suggested using their tangent lines. In our classroom. 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. r2 and the distance Classroom Problem 5: Construct two orthogonal circles in one plane. This is part of the need for communication. and any of these possibilities can occur by adjusting the parameters. relevant theorems. and so forth. The instructor asked “what if the radii are much greater B than OO ?” Homework: Determine the necessary and sufficient condition on the radii for the circles to meet. using perceptual evidence. they responded that the lines locally approximate the curves and we already know what the angle between lines means. If the circle and line have equations ( x a )2 ( y b)2 r 2 and y mx k . so algebra is more appropriate here. Circles centered at O and O intersect at A and B .9 between the line and any chosen line in the plane which meets it. The student made several further conjectures. OO AB and bisects AB . lines OO and AB are drawn and meet at C . potential counterexamples. A second student presented this reasoning. in that different meanings of the same term must be equivalent. . and it is advantageous to have multiple interpretations for a term. so that students progressively develop O C O' their skepticism and standards of proof. and this is equivalent to taking the complement of the angle between the line and the plane’s normal. the circles will be orthogonal if their radii are equal. a student presented her approach (see Figure 1). Once this equivalency has been established. there are two. the instructor may ask. A geometric approach to this depends on a careful axiomatization of geometry. She suggested that the radii must be greater than half OO in order for the circles to intersect. and specifically the idea of orthogonal curves. Since a quadratic equation results. Furthermore. Here the instructor can discuss the relationship between this meaning of “tangent line” and Euclid’s meaning as a line intersecting exactly once (which may be familiar to students from high-school). The angle between the line and its orthogonal projection into the plane will do. Asked why this makes sense. for otherwise it wouldn’t be possible to communicate meaningfully about the term. Calling the radii r1 . This is a proper answer. which the class examined critically: Figure 1: Constructing two orthogonal circles All four triangles such as OAC are congruent (in fact there are two congruent pairs). Such conjectures should always A be examined carefully. “Imagine that I am blind. How can I prove that given a circle. 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. The case with no intersections answers the given question. which is common among students. To define the angle between curves. there is a line that does not intersect it?” The “game” of playing blind is to remove the use of visual perception. Since the equation of the circle presents it as a level set of a function f ( x. say at a y-intercept x 0 . 2) . An algebraic approach may lead to more insight than calculus for many students. y ) . Draw one circle with any center O and radius r1 . The equation of the circle is ( x 1)2 ( y 2)2 16 . one can also find a normal vector by taking the gradient. draw a perpendicular segment AO of any length r2 . The two signs again reflect the symmetry of the circle about its diameters. the reasoning is valuable after we see how it fits into the structure of the problem. which look plausible from a picture. the construction can be made as follows. we need to tell algebra or geometry what a tangent line is. the negative Classroom Problem 6: A circle of radius 4 is centered at (1. 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. that is. We seek a line y mx b intersecting the circle only once. The slopes are 1 / 15 at the y -intercepts and 3 at the x -intercepts. for circles one can use the simplest definition: a line meeting the circle in exactly one point. They are of course symmetric about the vertical and horizontal diameters of the circle. As we saw. and then m 1 / (b 2) 1 / 15 .0) . if the radii are perpendicular there. students should have no trouble giving examples of lines which meet a parabola in one point but are not tangent to it. the expected slope. 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. Where does it cross the coordinate axes? What lines are tangent to it at those points? reciprocal of the slope of the radius to ( x . Indeed. the circles will be orthogonal if each radius to A or B is tangent to the other circle. Thus. or lines tangent to a non-convex closed curve which meet it more than once. . This can be confirmed by students who know calculus: implicit differentiation of the equation of the circle gives dy / dx ( x 1) / ( y 2) . then OQ is greater than the radius OP by the Pythagorean theorem. If this has only one root x 0 . then two coefficients must vanish: b 2 4b 11 0 gives b 2 15 . 2 15) and (1 2 3. This happens if r12 r22 s 2 (by the converse to the Pythagorean theorem). 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. Solving the equations of the line and circle simultaneously leads to the quadratic (1 m 2 ) x 2 2(mb 2m 1) x (b2 4b 11) 0. which should not be surprising since the Pythagorean theorem is the basis for the equation of the circle. They also follow from applying the Pythagorean theorem to a diagram. from which the intercepts are easily found as (0. y ) . However. so Q is outside the circle. at the endpoint A of a radius. if O is the center and Q some other point on this line. which gives twice the radius vector. so that the line and the circle have the same y-intercept.10 between the centers s . To find the tangent lines. As Euclid shows. 3 Supplementary and Practice Problems The following problems should be seen as a continuation of the Classroom Problems. and h in the equation x 2 y 2 fx gy h 0 and convert the resulting equation into an equation of the form ( x a )2 ( y b)2 r 2 . then the Pythagorean Theorem implies r12 r22 s 2 . there are calculusbased approaches to this problem using the slopes of the tangent (or normal) lines to verify orthogonality.1. Problem 2. So the condition does guarantee that the circles intersect. Conversely.11 Orthogonality condition for circles. and h a negative integer. and students should independently make sure they can prove them. with radii r1 . g 2 / 2) . is equivalent to a simple algebraic condition on the coefficients in their x 2 y 2 f 2 x g 2 y h2 0 are equations. Students may not realize that the converse to the Pythagorean Theorem is distinct from the theorem itself and requires a distinct proof. two distinct implications to prove. g1 / 2) and ( f 2 / 2. 4 4 1 so the distance formula gives s 2 ( f12 2 f1 f 2 f 22 g12 2 g1 g 2 g 22 ). Find the equation of a circle centered at the origin that is tangent to the line 2 x 2 y 39 . if follows that there is a right triangle having sides r1 . Again convert the resulting equation into an equation of the form ( x a ) 2 ( y b) 2 r 2 . If the circles are orthogonal. and orthogonally. Now choose another set of values for f . R. which means that the centers of the circles are close enough for them to intersect. an idea that will recur (Repeated Reasoning!) often. we find their radii given by 1 1 r12 ( f12 g12 ) h1 . This is a beautiful example of the PGA WoT (way of thinking): a natural geometric condition. Completing the squares in the equations of the circles. Problem 3: Let P. orthogonality of circles. this has to be assumed. r2 . Of course. r2 and centers separated by s . g an irrational number. or does it only make them orthogonal assuming they intersect? Since the condition implies r12 r22 s 2 . Choose any values for f . Their centers are ( f1 / 2. to clarify the subtle logic in this problem: Does the given algebraic condition ensure that the circles intersect. and h . Substituting indeed 4 produces the claimed algebraic condition. and S be four distinct points in the plane such that no three . Q. Then the triangle inequality says r1 r2 s and | r1 r2 | s . whereupon the converse to the Pythagorean Theorem shows that the circles are orthogonal. it does not guarantee that the “squared radii” computed from the equations are actually positive. g . r22 ( f 22 g 22 ) h2 . g . It is useful to keep track of the results from geometry that are being used to support the reasoning throughout this module. This is also an opportunity for students to orthogonal if and only if realize that equivalence (if and only if) means there are f1 f 2 g1 g 2 2(h1 h2 ). but this time f should be a fraction. and s . A final question for students. Problem 1. if the condition holds we can reverse the reasoning and deduce r12 r22 s 2 . However. Classroom Problem 7: Prove that the intersecting circles x 2 y 2 f1 x g1 y h1 0 and 1. b. and prove that there are no others. consider the 4 tangents at the points of intersection (1 tangent at each of 2 points for each of 2 circles). Are the circles orthogonal at these points? Problem 5: Find all the common tangents to the circles x 2 ( y 2)2 1 and ( x 2) 2 y 2 1. How many circles can you draw through the points P. what .C. Problem 7: Given 2 intersecting circles in a plane. a. where the two spheres intersect at more than one point. R and S? Problem 4: Find the intersection points of the circles ( x 1) 2 ( y 1) 2 5 and x 2 ( y 1) 2 4 . Q and R? d. For the last case. find an algebraic condition which ensures the two spheres (i) do not intersect. b ) and radius r . d. (iii) intersect at more than one point. Problem 6: How many common tangents can there be for two circles in the plane? Exhibit all the possibilities. c ) and radius r ? c. How many circles can you draw through the points P and Q? c. What is the equation of a sphere with center ( a. Define sphere geometrically. How many circles can you draw through the points P.F) are on the same circle. (ii) intersect at exactly one point. only 6 of which are distinct. b.12 of them are collinear. a. Draw perpendicular lines from each circle center to each tangent.E.B. thus obtaining 6 points of intersection: 4 lines from each center yields 8 points. Given two equations of two spheres.D. How many circles can you draw through the point P? b. Problem 8: We have seen that an equation of the form ( x a ) 2 ( y b) 2 r 2 represents a circle in the plane with center ( a . Q. Show that all these points (A. 0. what curve did the particle complete? Answer the same 2 question for 4 seconds and for 1 second. What are the curves along which 2 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. In Problem 8. and C.sin 2t . y (t ). B. (cos . respectively: t (cos t . and what is the distance between their tips? At what times do the hands coincide? observed just for . the 5 2 . At 12:00 AM. for example. A. move in space. This motion is useful for physical modeling but irrelevant for geometric properties of the locus.sin t ) . 0) . 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. If the particle’s movement was observed for 2 seconds. but also the motion of a point along that locus in time. (0. Consider the two parametric curves: r1 (t ) (cos t )i (sin t ) j and r2 (t ) (cos t ) j (sin t )k .13 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? x 2 y 2 9 ( py x )( x p 3) 0 In our classroom we did not introduce parametric equations of the circle. z (t )) is often written as r (t ) x(t )i y (t ) j z (t )k . including the part about being planar. to verify that the given curves are circles. Problem 13: Three particles.sin 4t ) . Here algebra encodes more information than the geometry requires. What is the 9 particle’s position (in the x . Some of the following problems serve to do so. Problem 10: A particle moves along the line 2 x 3 y 2 0 . They also need to observe that parametric equations describe not only a locus. Where are they located t minutes after noon? What is the angle between them at that time. What is the particle’s y -coordinate at that time? The 2 4 particle’s velocity in the direction of the x -axis is meter/second.cos 2t . The positions of the three objects in space at time t seconds from the start of their movement are. students need to use the definition of a circle. 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 t .sin t ) . what curve did the particle complete? If the particle’s movement was particle’s x -coordinate is seconds.y plane) at any given time t ? What is the particle’s velocity along the y -axis? Problem 11: The position function r (t ) ( x (t ). and represent it algebraically in a Cartesian coordinate system. and algebraic realities. formalize a geometric definition capturing it. we may talk of a geometric way of thinking in this manner. Attending. One of its instantiations is the realization that when applying algebra. geometric. Algebraic way of thinking. what can be assumed or chosen freely. An equation can be rewritten in various forms which make certain properties more noticeable. for which the intuition is a kind of oval. Algebraic invariance way of thinking. 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.14 2. Classroom Problem 4: Find a Cartesian equation of an ellipse. We asked students: what other geometric objects might we study in this way? One student suggested an ellipse. when there is a need. Another student suggested a formal definition. one must “tell algebra” all the relevant geometric constraints. Deductive reasoning: Logical structure of proofs: what is given. Likewise. Unit 2: The Ellipse Focus Ways of Thinking and Ways of Understanding Referential symbolic way of thinking. The link between the forms is provided by something which remains invariant. one must “tell geometry” all the given conditions. 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. Classroom Problem 1: What is an ellipse? [See pedagogical discussion of this problem. Try to state explicitly what it proved. what is proved. and a number d F1F2 . in solving a geometry problem. Algebra as deduction: Every algebraic computation is a proof of something. F2 in a plane. to a geometry problem for example. PGA way of thinking: Attending to interrelationships between physical/perceptual. which the instructor rephrased: Fix two points F1 . This is a broad way of thinking. namely.] 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). Usefulness of completing the square. Classroom Problem 3: Show that any chord AB of an ellipse passing through the center is bisected by the center. An ellipse is the locus of points P in the plane Classroom Problem 1: What is an ellipse? . for example the solution set. to the meaning of symbols and their manipulations. Distinguishing between a theorem and its converse. is the locus of points whose distances from focus and directrix are in the ratio e 1 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. because A is inside F1BF2 . AO OB ? Then choose A on OB with OA OA (see Figure 3). the instructor knew the status of student conjectures but did not reveal this fact. After several unsuccessful attempts by students. A' the instructor suggested a proof by contradiction: what B disaster follows if. The conclusion would follow from TO a . The instructor added: sometimes the teacher is an actor. This problem serves to check that our formal definition can justify properties of the ellipse that we intuitively expect to hold. To prove AO OB . In this case. The instructor asked students: what is the pedagogical value of this proof? Students suggested: present facts as they are needed in the proof. F2 However. AF1 AF2 BF1 BF2 . sometimes he is genuinely thinking at the board. T c OF1 OF2 . Classroom Problem 2: Show that the major axis of an ellipse is greater than the minor axis. how do we tell geometry what is given? That A and B are on the ellipse is expressed by AF1 AF2 BF1 BF2 . also F1 that F2 AF1 is bisected by AB . extract subdiagrams from diagrams and draw them separately. Let T . A a vertex on the minor axis. This is contrary to intuition. whose proof the instructor reviewed. so the sum of its distances to the vertices F1 and F2 must be less than B 's. the need for indirect proof was unexpected. Students A conjectured that AF1 BF2 but could not justify it. that is. Define a AF1 AF2 so that d 2a . and a b .21. Part of the important lesson from this problem was that proofs are not done just to have a proof. which would follow from TF2 T F1 . Then AF1 AF2 2a. the steps must still be assembled into a complete and organized argument [do so!]. also TF1 TF2 2a. Call b OA and A a b T' F1 O c F2 Figure 2: Ellipse axes Classroom Problem 3: Show that any chord AB of an ellipse passing through the center is bisected by the center.15 such that PF1 PF2 d . the teacher is human and can demonstrate getting stuck and yet recovering. The instructor drew one on the board using a shoelace wrapped around two pegs (perceptual verification that the definition is compatible with our intuition). Another intuition about the ellipse to verify is that any chord AB through the center O should be bisected by the center (symmetry). and O the center (the midpoint of F1F2 ). AOF1 BOF2 since they are vertical angles. The intuition is confirmed by Euclid's Proposition I. which is not true. T be the vertices on the major axis with T closer to F2 . This follows from TF1 TF2 T F1 T F2 upon substituting TF2 2c for TF1 and T F1 2c for T F2 . Also. but also to learn new things along the way. O . contrary to expectation. from the definition of the center. Figure 3: Bisected ellipse chord From congruent triangles it now follows that both B and A are on the ellipse. OF1 OF2 . Knowing the sum and difference of the focal radii.2) r22 ( c x ) 2 y 2 . but for now putting the origin at the center and the foci at ( c.1). Now there is a Need for Computation to verify the conjecture by reducing (1. and gives the new equation ( x 2 / a 2 ) ( y 2 / b2 ) 1 . we can solve for each. For example. we translate our definition of ellipse into algebra. which a student had remembered.1) to this simpler form. visually. 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. so the ellipse is symmetric about its center. Students should not fear lengthy algebra. and in fact more: the ellipse is symmetric about both axes. (1. in the direction perpendicular to this axis. which we continued to promote throughout the module. (1. This reasoning is simpler and more insightful than the previous computation. determine c and hence the positions of the foci. the definition gives ( x c ) 2 y 2 ( x c ) 2 y 2 2a. A simpler alternative to the brute-force algebra is to introduce notation for the two focal radii and write equations r12 ( c x ) 2 y 2 .1) As c 0 this becomes the circle x 2 y 2 a 2 . Factoring and recalling r1 r2 2a gives r1 r2 2cx / a . obtaining the unexpectedly simple formulas cx r1 a . a (1. the ellipse seems to be a squashed version of this circle.2) produces the standard form of the equation of the ellipse.3) cx r2 a . It follows that the major axis is uniquely determined as the longest chord through the ellipse. The class therefore conjectured that this equation is equivalent to (1. does an ellipse determine its foci? The ellipse is obtained by squashing a circle whose diameter is the major axis. Classroom Problem 4: Find a Cartesian equation of an ellipse. squashing the y coordinate by a factor b / a should produce the correct minor axis. a Substituting back in (1. Their lengths. Indeed. and be alert to the possibility of meaning-based simplification. and the minor axis as its perpendicular bisector. but should pause often to unpack the meaning of the symbols and equations. The composition of these reflections is reflection about the center. we showed that a circle determines its center. Does it confirm the intuition that an ellipse can be obtained by “squashing” a circle? For this problem. Allowing students to see this promotes the Referential Symbolic WoT. We will later consider general position and orientation of the ellipse.16 Closely related to this problem is the claim that the ellipse is symmetric about its center. We can prove this. Reflection about the perpendicular bisector of the focal segment (minor axis) interchanges the two focal distances and so preserves their sum. (See Supplementary Problem 1 for a Euclidean construction . Subtracting gives the simple result r12 r22 4cx . The algebra is lengthy but straightforward and uses a 2 b2 c 2 . which is productive because we recognize the difference of two squares. 2a and 2b . The squashed circle viewpoint is very useful for PGA reasoning. 0) . and at the vertices of an ellipse. a variable point on the tangent line. and ( x . or the algebraic method of finding a line having a single intersection of multiplicity two with the ellipse. Let the angles between PF1. This can be done using calculus. The notational distinction between ( x1 . y1 ) in the first quadrant. and make angle with the major axis (see Figure 4). The result is ( x1 x / a 2 ) ( y1 y / b 2 ) 1 . deserves emphasis. the chosen point of tangency. tan y1 / (c x1 ). tell algebra that the line is tangent to the ellipse. providing some initial evidence for the conjecture. We want to prove that 1 2 . P F1 O F2 Figure 4: Characterization of the tangent line to an ellipse Students had several approaches to this problem. ( x1 . y ) . we need to determine the equation of the tangent line. Let Classroom Problem 5: We know that the tangent line to a circle is perpendicular to the radius at the point of tangency. that is. each promoting important Ways of Thinking. Let a line be tangent to an ellipse with equation in standard form at point P ( x1 . PF2 and the major axis be .17 of the foci. but then switched to the conjecture that these angles are equal.) In our classroom students briefly considered using the radius from the center. ( x . and the different approaches can be productively compared. tan y1 / ( x1 c). . y1 ) satisfies both the equation of the ellipse and that of the tangent line. y1 ) . but preferred the focal radii. First. y ) satisfies only the latter. their tangents) can be computed from the slopes of the lines: tan b 2 x1 / a 2 y1 . The Need for Communication requires distinguishing them. They conjectured that the angles between the tangent line and these may sum to 90 . using the same symbol for two objects would tell algebra that they are the same. respectively. Some angles in Figure 4 (more precisely. Students observed that this is true in the special case of a circle. Is there a similar characterization of the tangent line to an ellipse? the angles between the tangent line and the focal radii PFi be i . 0) and the required proportion becomes . here 0 . the terms b 2 x12 a 2 y12 in the numerators tan 2 simplify to a 2b 2 . Indeed. then by Euclid. 1 tan tan Here is an example of the Necessity for such trig identities (“When will I ever use this?”). it should have no units and thus the numerator and denominator must have the same overall power).18 We can use the exterior angle theorem to find 1 and 2 : 1 and 2 . so they are equal (note also that referential symbolic reasoning provides students with a cy1 check of their work: since the expression computed is a tangent. tan tan tan 1 . It suffices to show that the normal line at the point P of the ellipse bisects angle P in the triangle F1PF2 . y1[a 2 ( x1 c ) b2 x1 ] a 2 y12 b2 x1 ( c x1 ) . the point K has coordinates ( c 2 x1 / a 2 . for example. this is equivalent to showing that F1K / F2 K F1P / F2 P . VI. One finds straightforwardly (don't fear complex algebra!) b 2 x1 y 1 2 a y1 x1 c tan 1 . we need to consider whether tan 1 tan 2 suffices to conclude that 1 2 . multiplying one by the reciprocal of the other and simplifying). (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!] ( xa 2 / x1 ) ( yb 2 / y1 ) c 2 . b 2 x1 1 2 a ( x1 c) y1 b2 x 2 1 c x1 a y1 tan 2 . Each tangent then reduces to b2 . Prop. but the Referential Symbolic WoT suggests a meaning-based simplification. Finally. Then trig identities [prove them!] allow us to compute. This provides Necessity for the concept of a one-to-one function on a specific domain. y1[a 2 ( c x1 ) b2 x1 ] One can check the equality of these by brute force (e. and in the denominators a 2 b 2 c 2 can be used. from the equation of the ellipse.3. If this normal meets the opposite side F1 F2 at a point K . b 2 x1 1 2 a (c x1 ) We first clear denominators to obtain b2 x1 ( x1 c ) a 2 y12 tan 1 . A second proof of the equal-angles property of the tangent line requires less computation and provides more insight.g. and that reversing the algebraic reasoning may prove it. r2 e( f x ) . [We observe that f a . where we define e c / a and f a 2 / c . is the locus of points whose distances from focus and directrix are in the ratio e 1 The results obtained earlier for the focal radii. then we have an c x O F1 F2 alternative definition of an ellipse.19 c ( c 2 x1 / a 2 ) r1 . Our computation of r2 proved that. y ) to a vertical line. so that this line lies to the right of every point on the ellipse. Besides the solution following from the equal-angles property. P Most students do not yet possess the WoT that the converse should automatically be investigated. One way to establish this 2 is to begin with r2 ac ( ac x ) and apply the Pythagorean Theorem (twice) to Figure 5 to obtain Figure 5: Focus/directrix definition of ellipse . Finally. Begin with the circle x 2 y 2 a 2 and its tangent line at some point ( x1 . c ( c 2 x1 / a 2 ) r2 which follows from our earlier computation of the focal radii. e. are surprisingly simple--simple enough that the PGA WoT suggests seeking a geometric interpretation. But the rate of change of the distance to a focus is simply (minus) the component of velocity in the direction of the focus. if P lies on the ellipse. the sum of the rates of change of the focal distances is zero. r2 a ( cx / a ) . physicsbased proof which provides insight. Imagine an object moving around the ellipse. not necessarily constant. If so. That is. A linear function of x can be interpreted in terms of the distance from ( x . y ) to the vertical line l with equation x f . there is another following from the squashed circle viewpoint.] A useful Way of Thinking is that every algebraic computation is a proof of something. (1. and the squashing does not change the x -intercept of the tangent line. That is. and the tangent line to the ellipse is drawn by joining the intercept to P . r1 r2 Here we ask whether a point satisfying PF2 ePl is necessarily on the ellipse.g. and also that e 1 . In this case. again implying 1 2 . This says that v cos 1 v cos 2 0 . There is a third. with any velocity v . y1 ) . the focus/directrix definition. Squashing by a factor b / a in the vertical direction produces the ellipse ( x / a )2 ( y / b)2 1 along with its tangent line at the corresponding point P( x1 . and one should try to state explicitly what it proved. Classroom Problem 6: Is the focus/directrix definition of an ellipse equivalent to the two-focus definition? That is.3). by1 / a ) . from the tangent line to the circle we obtain this intercept. the directrix. perform a Euclidean construction of the tangent line at a specified point. a useful follow-up question is: given an ellipse with its foci. namely drawing the focal radii and bisecting the angle they form. r2 is e times the distance from ( x . Since the sum of its distances to the foci remains constant. the distance from P to the focus F2 is e (the eccentricity) times the distance to l . then PF2 ePl . Therefore. we need deductive reasoning to correct our intuition in this case. A plane meets the cone in the curve E (which we assume is closed). some are fixed.P.20 r12 r22 ( c x ) 2 ( x c ) 2 ( a ac x ) 2 . and some are variable like P . Intuition tells us that a “horizontal” plane cuts a cone in a circle. Thus. such as the cone and plane. It is also possible. In 1822. to take an algebraic approach: obtain equations for a cone and a plane cutting it in three-dimensional Cartesian coordinates. the focus/directrix property implies the two-focus definition of the ellipse. Since a ac x 0 for every point on the ellipse [prove it!]. In addition to the complexity of the algebra. Students need practice (Repeated Reasoning) with the overall logical structure of such proofs. lines. Our objective is to show that E is an ellipse. and cuts the plane in the line d . like the point of Figure 6: Dandelin sphere construction of ellipse tangency F . a cone has vertex V and is tangent to a sphere along circle C lying in plane and having center O . P is an arbitrary point on E . it seems unlikely that this oval will be symmetric about its center rather than egg-shaped. for any plane cutting the cone in a closed curve E . Perceptually. It is important for students to attend to the temporal (logical) sequence in which elements are added to the diagram: which points. In Figure 6. is tangent to the sphere at F . The cutting angle is PDM and the cone's elevation angle is VLM . Some elements are given. In our classroom we motivated this construction and presented the reasoning in the case of the ellipse. G. and add to obtain r1 r2 2a . the generator VP meets at L . and so forth are chosen first and which others are then determined by these choices. and derive an equation for their intersection. whereas if the plane is tilted slightly the intersection is an oval curve which may be an ellipse. y are not also standard Cartesian coordinates in the cutting plane. Classroom Problem 7: Is a (suitable) plane section of a cone an ellipse? . we can take the square root to obtain r1 a ac x . we can draw a sphere tangent to both the cone and the plane as shown. and PM is drawn perpendicular to . PD is drawn perpendicular to d . 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. Conversely. where we want to have an equation for the locus. but complicated. a conceptual difficulty is that the three-dimensional Cartesian coordinates x. Dandelin discovered a beautiful proof that “conic sections” are indeed plane sections of a cone. ) We also have PM / PL PM / PF sin and PM / PD sin . (Students are familiar with the fact that tangents to a circle from an external point are equal. Connect the midpoints M 1 and M 2 . as the plane becomes horizontal. begin with a circle and a point other than its center on the line through its center and perpendicular to its plane. we can find the midpoint without . Supplementary and Practice Problems: Problem 1. Construct a second pair of parallel chords EF and GH not parallel to the first pair. 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. By way of telling geometry that the sphere is inscribed in the cone. We are relying on this intuitive topology for the conclusion that the curve is a complete ellipse. Then the endpoints of the first chord are found by solving b2 m 2 a 2 x 2 2mn1a 2 x (n12 a 2b2 ) 0 . but we returned to it when comparing the Dandelin sphere proofs for the different conic sections.21 To prove that E is an ellipse. and connect their midpoints M 3 and M 4 . take the union of all lines joining this point to the points of the circle. but had difficulty deducing the corresponding fact for spheres. we need to identify candidates for the focus and directrix. Let the ellipse be given by ( x / a )2 ( y / b)2 1 . Students do not always realize initially that this defines a double cone. it also raises several subtle issues which should be made explicit if not necessarily resolved rigorously in class. Then the intersection of M 1M 2 and M 3 M 4 is the center of the ellipse. Although the construction is purely synthetic. To obtain the latter from the former. To prove this. and d might be the directrix. (Indeed. the fact that a cutting plane with produces a closed curve of intersection relies on some intuitive topology. Although Dandelin's proof is visual and elegant. Second. Therefore. the proof uses coordinates. Some support for this conjecture comes from the facts that. That in turn requires . PD sin a constant independent of P . If we know that e 1 . 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. For example. Construct a pair of parallel chords (AB and CD in Figure 6). it suffices to show that the center is contained on the line joining midpoints of any 2 parallel chords. F moves to the center of the circle E and d moves off to infinity. the definition of a cone as a figure of solid geometry may still be at the intuitive level for students and should be formalized. which implies e 1 . the proof shows only that the curve of intersection is contained in an ellipse. PF sin e. This follows from our implicit assumption that the plane cuts the cone in a closed curve.) We did not press the issue of what it means to be a closed curve at this point. we find the center. we note that PL PF are tangents to the sphere from the same external point. However. First. The point of tangency F might be the focus. Given an ellipse. which occurs only when the plane cuts each generator of the cone. one has to find the right circular cross-section of the sphere. then we have shown that E is an ellipse. First. Third. construct (synthetically) its foci. Suppose one chord is given by the line y mx n1 and the parallel chord is given by y mx n2 . The circle will intersect the G ellipse at 4 points. 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. Problem 5. which we connect to get a M3 rectangle whose sides are parallel to the B H desired axes [prove this!]. Let E be an ellipse with center O . Find the eccentricity and the directrices of the ellipse 2 2 x 2 y 3 1. and minor axis of length 2b .22 mn1a 2 (average of 2 solutions to a b2 m 2a 2 m 2n1a 2 n1 . Use a second Dandelin sphere to confirm that an ellipse has a second focus/directrix pair. Figure 7: Constructing ellipse foci 1 With the axes in hand. y2 ) 2 . The intersection points of this circle with the major axis are the foci. pick a point P on the ellipse that is also on the minor axis. 5 4 Problem 6. We can then draw D lines parallel to these sides through the origin. 2 n2 . Similarly. Problem 2. To find the axes. we verify that y1 y2 . x1 x2 yi m 2 ni a 2 ni (b 2 m 2 a 2 ) xi mni a 2 m 2 a 2 (b 2 m 2 a 2 ) A ma 2 E C b2 M4 . Prove that if P E . then b PO a .0 and directrix the line x 4. giving the desired M1 equality. major axis of length 2a . so y-coordinate is y1 2 b m 2a 2 mn2a 2 m 2 n2 a 2 be ( x2 . Find the equation of the ellipse with foci 23 . Problem 4. the other midpoint will quadratic is B / 2 A) . we pick any point on O the ellipse and draw a circle centered at O M2 through that point. Problem 3. Consider the equations . Draw a circle centered at P with radius equal to the semimajor axis.0 and 23 . ma 2 F which is independent of ni . To show that the line connecting the midpoints 2 2 2 2 b m a b m a finding the endpoint: its x-coordinate will be x1 contains the origin. Analogous statements for the parabola and hyperbola can be proved similarly. one might assume that the ellipse's axes are parallel to the sides of the square. In the simplest approach to the problem. a larger ellipse will fit diagonally. Find an equation for an ellipse with the same foci as the ellipse 25 16 1 but greater eccentricity. so the constraint is simply that the major axis 2 r / cos not exceed the side 4r of the square. Problem 12. a. Is a plane section of a circular cylinder also an ellipse? Perceptually. b. Find a normal line to the ellipse 3x 2 2 y 2 50 which is perpendicular to the line 6x y 7 . A mining company drills a mineshaft into the ground. The sides of the square are then tangent to the ellipse and make a 45 degree angle with the axes. Let PQ be a diameter of an ellipse. Prove that if PQ is a diameter of an ellipse. where r is the radius of the hole. Again. then the tangent lines to the ellipse at the points P and Q are parallel. However. Problem 11. The Dandelin sphere argument actually applies to the cylinder as well as the cone. or / 3 . If the (elliptical) opening at the surface must fit within the square plot of ground owned by the company. Graph both ellipses on the same set of axes. rather than squashing. the coordinates of the foci. this is an excellent opportunity for students to appreciate the advantages of the Deductive over the Empirical proof scheme. This gives cos 1 / 2 . Find a tangent line to the ellipse x 2 4 y 2 196 which is perpendicular to the line x 5. Determining where such lines are tangent is a nontrivial exercise. However. and the length of the minor axis. with its axis making an angle with the vertical. how does this restrict ? If the axes of the ellipse are assumed parallel to the sides of the square. Show that one of the above equations defines an ellipse and that the other equation does not define an ellipse. Problem 8. Then the elliptical opening can be described by the equation x 2 cos2 y 2 r 2 . 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 . we see the cylindrical mineshaft as if its axis were vertical. Let P E . For the equation which does define an ellipse. Rather . find the coordinates of the center. b. Prove that the midpoints of the chords of the ellipse which are parallel to PQ all lie on the same line. y2 x2 Problem 7. it seems unlikely that sections of cylinders and cones have the same shape. this is an easy application of the squashing of the circle to an ellipse. the length of the major axis. The hole is a circular cylinder.23 4 x 2 5 y 2 16 x 10 y 39 0 4 x 2 5 y 2 16 x 10 y 39 0. Let E be an ellipse with center O . If we tilt our heads by the angle . Problem 13. Problem 9. a circle. and the ground as a cutting plane with elevation angle . a. Here the ellipse is obtained by stretching. a larger ellipse will fit in the square with the major axis along the diagonal. Problem 10. whose side is twice the diameter of the cylinder. Find an expression for PO in terms of the angle between PO and the major axis of E . Problem 14. which in our case gives a distance of (16 89 ) / 2 4. If a straight river runs through the field along a given line. As in Problem ellipse9. so its side is this value times 2 . and the rope has length l .37796. while the river has equation y x 16 . and the distance between this line and the river is the closest the dog can get. The minus sign gives the tangent line nearer to the river. c 3. (Of course. To achieve this we use implicit differentiation to locate the points where the ellipse has slope 1 . Problem 15. . This translates into 67. The problem constraint is then r 2 1 cos2 4r. If the posts are separated by a distance d . must then be tangent to the ellipse. other methods of finding the tangent lines. cos2 ). how close to this river can the dog get? Consider a numerical example in which l 20 and d 6 . finding 2 x cos2 2 yy 0 with y 1 . Let the ellipse be centered at the origin with major axis along the x -axis. cos which can be solved to yield simply cos 1 / 7 0. with slopes 1 . it may be easier to draw a tilted square around the standard ellipse. this restricts the dog to an elliptical region in which the sum of focal distances is less than l d .24 than inscribe a tilted ellipse inside the square.) Substituting back into the equation of the ellipse gives the points of tangency as r ( x. cos The four tangent lines form a square whose diagonal is twice this value. there are two tangent lines to the ellipse parallel to the river. versus 600 as found previously when the major axis was parallel to the sides. The x -intercept of the tangent line at the point with positive coordinates can be computed as r 1 cos2 .79 o . namely y x 89 . The sides of the square. and the equation of the ellipse is ( x 2 / 49) ( y 2 / 40) 1 . so that y x cos2 . y ) ( 1. 2 cos 1 cos where all four sign combinations are possible. lead to the same result. The distance between parallel lines of slope m whose y -intercepts differ by b is b / m 2 1 .6429 . discussed above. Then a 7. A dog is tied up in a large field by a loop of rope running through its collar and around two posts. Structural way of thinking. choose a point Q on l .0) . Usefulness of completing the square. directrix Ax By C 0 . but students should recognize that reversing the roles of x and y all but the last to be put into this standard form. An equation can be rewritten in various forms which make certain properties more noticeable. PGA way of thinking: Attending to interrelationships between physical/perceptual. and that algebraic transformations induce a translation of the locus. and algebraic realities. Algebraic invariance way of thinking. Then the . and they presented a geometric construction of points on the locus. The link between the forms is provided by something which remains invariant. Students should see a need to reconcile these equations with their previous understanding of a parabola. draw FQ and mark its midpoint M . what happens for e 1 ?). directrix y -axis: y 2 2kx k 2 . Try to generalize established properties and explore a parameter space (if an ellipse has 0 e 1 . directrix x k : y 2 4kx . directrix x k / 2 : y 2 2kx . how can the other equations be transformed into this one? Pedagogical Considerations: Classroom Problem 1: What locus satisfies the focus/directrix definition with e 1 ? 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.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. Examples that were discussed in our class include: Focus ( k . Classroom Problem 1 What locus satisfies the focus/directrix definition with e 1 ? 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. Focus (k / 2.0) . for example the solution set. assuming without loss of generality that A2 B 2 1 : B 2 x 2 A2 y 2 2 ACx 2 BCy 2 ABxy C 2 . with a general line as directrix. geometric. Focus (0. 0) . probably defined as the graph of a quadratic function y ax 2 bx c . Focus (0. directrix x -axis: x 2 2ky k 2 . Given the focus F and directrix l . creates Necessity for the formula for the distance from a point (focus) to a line (see line unit). k ) .0) . Only one of the above equations has this form. One pair of students in our classroom interpreted this problem geometrically rather than algebraically. The last example. Focus ( k . or y ( x 2 k 2 ) / 2k . Coincidentally. the new curve has equation f ( x a. Does it mean that each point P ( x. y ) f ( x a . if we move each point as stated. k ) and l to be the x -axis. and shows.26 F M P l Q Figure 8: Geometric construction of points on parabola perpendicular bisector of FQ and the perpendicular to l at Q meet at a point P which is on the locus (see Figure 8). this problem is so simple that the parametric representation immediately reduces to the Cartesian version. one of the previously calculated equations for a parablola. so that f ( x. how can the other equations be transformed into this one? . y ) . 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 .0) . y ) 0 in order. 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. 2 0 2 The intent of this problem is for students to develop the technique of translating axes to simplify the equation of a curve. the coordinate of the arbitrary point Q . k / 2) and P is at ( x. One possibility is to introduce x x a. One easily calculates that M is ( x0 / 2. It also provides Necessity for confronting the distinction between a Cartesian equation and a parametric representation of a locus. or are we moving the coordinate axes while keeping the same locus? In our classroom we chose the former. First. Repeated reasoning will be needed for students to understand the source of the sign change. greater understanding of the construction and the definition of parabola. parametrized by x0 . but we postpone the use of rotations of axes until later. How do we modify the equation f ( x. The construction naturally gives a parametric representation. y ) on the curve is moved to the new location P( x a . say.) Students must carefully reason through the process to avoid subtle sign errors. it is necessary to decide whether the translations will be active or passive: are we moving the locus on a fixed coordinate system. Taking F to be (0. y ). or does it mean that we modify the equation by substituting x a for each occurrence of x ? In fact. y ) where x x0 and y ( x k ) / 2k . One can use this construction to see that the locus is symmetric about an axis through F and normal to l . The required condition FP QP follows from FMP QMP . A student suggests that it requires. by SAS. y ) 0 . which seems easier. Eliminating x0 yields y ( x 2 k 2 ) / 2k . 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. 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. y y for the coordinates of the moved point. let Q be at ( x0 . (They can also interchange x and y to reflect about the line y x . Another student translated this construction directly into algebra. y ) 0 . Problem 4: Give an example of an ellipse and a parabola with the same focus and directrix. 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 ? e. with equal probability of landing anywhere on it. Rotations by 90 degrees give the parabolas corresponding to the other sides. Problem 3: Find the focus and the directrix of the parabola y 2 x 2 16 x 31 . Let C be the set of points in the plane satisfying the equation f ( x. Problem 6: Prove that an appropriate intersection of a plane and cone gives a parabola. Unit 4: The Hyperbola Focus WoT's and WoU's Determining tangent lines and asymptotes. Problem 2: Find the focus and the directrix of the parabola y 2 x 2 . the parabola y (1 x 2 ) / 2 . Let C be the set of points in the plane satisfying the equation f ( x. Referential symbolic reasoning: “simplifying” the equation of a locus by translation of axes. The probability is obtained by dividing by 4 . The locus of points equidistant from the center and the side y 1 is the parabola having these as focus and directrix. Let C be the set of points obtained by translating the points in C down 3 units and to the right 5 units. 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. Using the octagonal symmetry. so that the center is at the origin and the area is 4 . PGA WoT. the area is 8 times that of the segment in the first quadrant bounded by the y -axis. Let C be the set of points in the plane satisfying the equation f ( x. y ) 0 . and is 2 1 0 (1 2 x x 2 )dx (4 2 5) / 3 0. What is the probability that it lands closer to the center than to any edge? Take the square to be bounded by the lines x 1 and y 1 . 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.219. the area of the square. and the line y x . its equation is y (1 x 2 ) / 2 . roughly an octagon with curved sides.27 Supplementary and Practice Problems: Problem 1: c. What equation is satisfied by the points in C ? f. Find an equation defining the parabola obtained by translating the parabola y 2 2kx 0 down m units and to the left n units. . students should also realize that the double Dandelin sphere argument does not apply. 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. Problem 8: (Putnam Exam) A dart is thrown at a square dartboard. y ) 0 . The dart lands closer to the center than to any edge if it lands in the closed region bounded by these parabolas. Problem 5: Find a tangent line to the curve 5 y 2 x 0 that is perpendicular to the line 20 x y 1 . and completing the square in x leads to k k 2e 2 ( e 2 1)( x 2 ) 2 y 2 2 . collecting terms. what happens for e 1 ?). Expanding. y (b / a ) x 2 a 2 .28 Structural way of thinking. n so that the line y mx n is tangent to the hyperbola ( x 2 / a 2 ) ( y 2 / b2 ) 1 . b. Then the defining property PF e Pl immediately gives Classroom Problem 1: What locus satisfies the focus/directrix definition with e 1 ? ( x k )2 y 2 e2 x 2 . a2 b This tells us that there is a better choice of coordinates than the one we made: we should transform x x e2k1 and y y . The equation gets more complicated at each step. The final form of the equation of the hyperbola is indeed ( x 2 / a 2 ) ( y 2 / b2 ) 1 . Try to generalize established properties and explore a parameter space (if an ellipse has 0 e 1 . e 1 e 1 2 2 2 2 2 2 2 2 2 Introducing a k e / (e 1) and b k e / (e 1) (since both quantities are positive) leads to ( x e2k1 ) 2 y 2 2 1. Its x -intercepts are ( a . like x ? Can we prove this? What does it mean? A .0) and the directrix l is along the y -axis. 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. y ) on the curve should be moved right to the location given by the shifted coordinates. To verify this. We now wish to “simplify” this equation to a “symmetric” form like that of the ellipse. m. and even the final form is about as long as the initial form. The following computation is remarkable in that no high school student would recognize it as simplification. 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. (b) Prove that the tangent line to a hyperbola bisects the angle between the focal radii. students should pose this question themselves.0) . Classroom Problems Classroom Problem 1: What locus satisfies the focus/directrix definition with e 1 ? Classroom Problem 2: What are the asymptotes of a hyperbola? Does a parabola have asymptotes? Classroom Problem 3: (a) Find a condition on a. The arrows mean that each point ( x. for example making any symmetry of the locus obvious. meaning a curve with equation ( x 2 / a 2 ) ( y 2 / b2 ) 1 . Based on their experience. Usefulness of completing the square. our students conjectured that the locus is a hyperbola. Pedagogical Considerations By this point. Perhaps it is monotonically increasing. (c) Find the angle between a hyperbola and an ellipse that share the same foci. students suggested a coordinate system in which the focus F is at ( k . Classroom Problem 2: What are the asymptotes of a hyperbola? Does a parabola have asymptotes? 4 P 2 K -5 F1 F2 -2 5 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 textbook derivation employs 20/20 hindsight to situate the focus at ( ae. and the asymptotes are these lines through the origin (center). Therefore. From the equation of the hyperbola. y1 b Thus. one obtains x1 a 1 (b / y1 ) 2 . The line tangent to the hyperbola at ( x1 . One can show also that the vertical (or . there is a second focus at ( ae. 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. as y1 . Note that a / e a ae . This is true because the composition of the increasing functions x 2 a 2 and x is increasing.29 student suggested that larger x 's give larger y 's. calculus is not needed here. which have been lost in translation (bad pun). 0) . (b / a ) x12 a 2 (b / a ) x22 a 2 . The x x1 intercept clearly approaches the origin as our point Figure 9: Limiting behavior of tangent line to a moves to infinity. Since the shift right was by a / e . 0) and the directrix at x a / e from the beginning. 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 directrix is now the line x a / e .4) has the limiting form y (b / a ) x . We need to locate the focus and directrix of the hyperbola given by our final equation. How can we tell algebra this meaning? One possibility is that if x1 x2 . the tangent line (1.then y ( x1 ) y ( x2 ) . We contrasted this with our implementation of the Necessity Principle and the PGA WoT. Using the derivative to test for an increasing function would be overkill. y1 ) has the equation xx1 yy1 2 1. Because of the symmetry of the curve. (1.4) a2 b This can be rewritten b2 x a2 y 2 1 x . miraculously resulting in the simple standard form of the equation. 0 . we must have x1 also. the form of the slope necessitates x1 hyperbola examining the ratio y1 . a y1 y1 a2 showing the slope and x-intercept K . that is. in such a way that x1 / y1 has the limiting value a / b . 0) and a corresponding directrix at x a / e . Applying the same shift to the focus locates it at ( ae. (c) Find the angle between a hyperbola and an ellipse that share the same foci. They should be related to the asymptotes. if an ellipse and a hyperbola with the same foci intersect at P . b. (It also has no center. y1 2 In the limit y1 . take a plane cutting a cone in the hyperbola.) The tangent line to the parabola y 2 2kx at ( x1 . To see the three-dimensional origin of the asymptotes. then the tangent line to the hyperbola at P coincides with the normal to the ellipse there. and a parallel plane through the vertex cutting it in a pair of lines. y1 ) has the equation yy1 k ( x x1 ) . which can also be written k y y x 1. For part (c). but unfortunately do not lie in the plane of the hyperbola. one more parameter such as the eccentricity is required. Classroom Problem 3: (a) Find a condition on a. Thus we have a whole family of ellipses and hyperbolas.0) and directrices the lines . the slope goes to zero and the intercept to infinity. m. Therefore the two curves are orthogonal at P .0) and (5. A parallel treatment of the parabola shows that it has no asymptotes. n so that the line y mx n is tangent to the hyperbola ( x 2 / a 2 ) ( y 2 / b2 ) 1 . The tangency condition in part (a) is a 2 m 2 n 2 b 2 . Students may not realize without reflection that neither the ellipse nor the hyperbola is completely determined by specifying their foci. and translating them to pass through the center. The equal-angle property of the hyperbola can be established in any of the ways used for the ellipse. indeed gives the asymptotes. because both bisect the same angle between the focal radii (See Figure 10). The hyperbola meets every generator of the cone except for these two lines. 8 6 4 2 P K -5 F1 F2 5 -2 This type of problem should be straightforward by now. the tangent line has no limiting position. (b) Prove that the tangent line to a hyperbola bisects the angle between the focal radii. and all are orthogonal when they intersect! -4 -6 Figure 10: Ellipse and hyperbola with the same foci Supplementary and Practice Problems: Problem 1: Find the equation of the hyperbola with foci ( 5. Projecting them into this plane.30 perpendicular) distance from the hyperbola to an asymptote goes to zero in the limit. where O is the center of the hyperbola. B . Reconstruct their method. eccentricity. Problem 2: Find the center. (a) Show that P is the midpoint of the segment AB . 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. (a) Given a line segment AB . Problem 4: Prove that an appropriate intersection of a plane and cone gives a hyperbola. . draw a circle with center O passing through A and B . and asymptotes of the hyperbola x 2 2 x 4 y 2 3 0. directrices. and eccentricity. Show that P is a point on the hyperbola if and only if | PF1 PF2 | 2a. Next draw the hyperbola determined as above by segment AB . Problem 6: The Greeks (Pappus) used conic sections to trisect angles. directrix. show that the locus of all points P such that PBA 2PAB is (a branch of) a hyperbola. Assuming without loss of generality that OA OB . Identify its focus. (b) Show that the area of the triangle ABO .31 x 16 / 5 . (b) An angle AOB can be trisected as follows. is independent of which point P was chosen. Show that 3POB AOB . Problem 3: Let F1 and F2 be the foci of the hyperbola x2 a2 y2 b2 1 . Let it cut the circle at a point P in the interior of the angle. foci. as follows. which we denote by particular coordinates. call them P( x1 . A third approach (requiring some familiarity with the notion of betweenness. respectively. y2 ) . y1 ) and ( x2 . where b y2 mx2 . QRT and PQS are corresponding angles formed by the parallel lines RT and QS and the transversal l and are therefore congruent. y ) be any other point on the line l . QTR is a right angle. B 2 y2 2 y1 . PSQ is a right angle. y ) is a point on l . y ) on the line satisfies the equation y mx b . Moreover. A second approach is to suppose that the given line l is the perpendicular bisector of some segment PQ . we may assume that x1 x2 x . respectively. Since RT is parallel to the y axis and QT is parallel to the x axis. y1 ) Classroom Problem 1: Find the equation of a line. both perceptually and mathematically. y2 ) . y2 ) . we have ( x x1 ) 2 ( y y1 ) 2 ( x x2 ) 2 ( y y2 ) 2 . and C x12 x22 y12 y22 . Similarly. the preceding processes of characterizing various conic sections should have prepared these students to recognize that the following problems are indeed legitimate questions. However. Let R ( x. P~R~Q . or equivalently. Students should be able to explain why generalizing this argument implies that any point ( x. The first approach uses similar triangles. we find that y mx b . 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. It follows that QS PS RT QT . we will choose this system in such a way that l is not parallel to either of the coordinate axes. we do not place a unit on the line at the beginning of this module—students would not understand the questions being asked at that point. Pedagogical Considerations Here we list three approaches to the problem. Ax By C 0 . Thus QRT is similar to PQS by AA similarity. and ( x. Let S and T be the points with coordinates ( x2 . x2 x1 x x2 Naming this common ratio m and solving for y . In order to tell algebra about the line l . y ) . Classroom Problem 1: Find the equation of a line. Without loss of generality.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. They should also realize the reason for fixing two points that define the line. Consider a line l . which we denote with general coordinates ( x. that is: y2 y1 y y2 . then since ( x. then the equation Ax By C 0 defines a line. and considering any other point on the line. If the coordinates of P and Q are ( x1 . A line is determined by two points. we should choose a coordinate system. y ) is equidistant from P and Q . y1 ) and ( x. where A 2 x2 2 x1 . and Q( x2 . Classroom Problem 2: Show that if A and B are both non-zero. one must show that the measure of PQR is . y2 ) . Prove these two assertions. For the second approach. in which case it is clear that R has the desired properties iff it has y-coordinate of 0. and R( x3 . the calculation is lengthy. k B ) where k A B2 B 2 C . Classroom Problem 2: Show that if A and B are both nonzero. Q ( x2 .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. Note that the cases of horizontal and vertical lines ( Ax By C 0 where exactly one of the coefficients A and B is zero) can then be treated as a composition of a rotation and a translation. That is. y2 ) . consider the segment whose endpoints have coordinates (0. if and only if Ax By C 0 . However. {R : Q ~ P ~ R & QP PR QR} The necessity for all 3 of these sets can be established through overaching problem 8. then the equation Ax By C 0 defines a line. {R : R ~ Q ~ P & RQ QP RP} 2. we can apply translation and rotation to obtain the general form of any line.0). That is. then Q is between P and R ). 2. we define a primary line as the set of all ordered pairs (x. 2. y3 ) satisfy the equation Ax By C 0 . From an expert perspective. Supplementary and Practice Problems: Problem 1: In school you learned that a line in the plane has the equation y mx b . suppose that the points P( x1 . one may introduce points S ( x2 . Then ( x. y ) is on the perpendicular bisector of this segment if and only if ( x 2 ( y k ) 2 ( x A) 2 ( y k B ) 2 . All of the approaches to the previous problem can be reversed. From this. y1 ) and T ( x3 . {R : Q ~ R ~ P & RQ RP QP} 3. The conclusion then follows since PQS and RQT are complementary and SQT is a right angle by construction. we define PQ as the union of 3 sets: 1. if P . we can choose (say) the x-axis to be along the direction of PQ . As above. where x1 x2 x3 .y plane has the equation y mx b . y1 ) . or y=0. For the similar triangles approach. A line in the x . explain the meaning of the following two assertions: a. If one blindly applies the distance formula. What does this mean? Specifically. If we assume that ( xi . where the equality of the ratios RT QT QS PS and follows from the equations Axi Byi C 0 for i 1. b. In order to show that PR is a line. then the result follows from a straightforward calculation. The third approach relies on the converse of the Betweenness Theorem (namely. . for i 1.3 and that the points satisfy the hypotheses of the theorem. Q and R are distinct points such that PQ + QR = PR . k ) 2 2 and ( A. The equation y mx b represents a line. this time using SAS similarity to conclude that QRT is similar to PQS . this is really defining a line through defining an axis. or equivalently.3 . given points P and Q. yi ) satisfy the equation Ax By C 0 . 34 Problem 2: Given two points in the plane. b ) and ( c. prove that the equation of the line yb xa . y0 ) to the line Ax By C 0 is given | Ax0 By0 C | by . d ) . that goes through these points is d b ca Problem 3: Show that the distance from the point ( x0 . ( a . A2 B 2 . has the equation this problem to the special case A B in which B 2 x 2 A2 y 2 2 ACx 2 BCy 2 ABxy C 2 . An appropriate capstone to this module would be the use of rotations to classify the conics represented by general quadratic equations in two variables. Algebraic invariance WoT. a student investigates the effect of other simple (linear) transformations on the circle. and a clockwise rotation by angle parabola with focus at the origin and with tan B / A will make the directrix directrix the line Ax By C 0. Ideally. we were only able to take the first step in this direction. The new coordinates are r (cos cos sin sin . Necessitating the concept of eigenvector by locating the vertices of a rotated ellipse. A convenient This should simplify to one of our standard way to implement the rotation algebraically is forms by rotating the parabola so that its via polar coordinates. B ) is normal to the Classroom Problem 1: Recall that a directrix. Need for Structure: classifying the conic sections represented by general quadratic equations.35 Unit 6: Conic Sections and Plane Rotations Focus WoT's and WoU's PGA WoT. with vertical. we would like to connect the discussion to quadratic forms and the eigenvalues and eigenvectors of linear transformations. the vector ( A. In fact. with A2 B 2 1 . Classroom Problem 1: Recall that a parabola with focus at the origin and directrix the line Ax By C 0. has the equation B 2 x 2 A2 y 2 2 ACx 2 BCy 2 ABxy C 2 . ) . Because our students had limited linear algebra background. we limited A2 B 2 1 . y ) on the circle. How is such a rotation carried out algebraically? Classroom Problem 2: Intrigued by the idea of obtaining an ellipse by squashing a circle. y ) has polar directrix is vertical. in our classroom. Effect of placement of locus relative to coordinate system on equation of locus. Beginning with the circle x 2 y 2 r 2 . or . Expressing this in terms of the original Cartesian coordinates necessitates the angle sum/difference identities from trigonometry (“When will I ever use this?”). the needed rotation angle is 45 . algebraic transformation inducing a rotation of the locus. which rotates to ( r. A point ( x. What curve does she obtain? What equation could she have graphed to obtain the new curve directly? Classroom Problem 3: What curve is represented by a general equation Ax 2 Bxy Cy 2 F 0 ? By a more general equation Ax 2 Bxy Cy 2 Dx Ey F 0 ? Describe the curves geometrically as completely as possible. sin cos cos sin ). This should simplify to one of our standard forms by rotating the parabola so that its directrix is vertical. she creates a new curve by plotting the new points ( xˆ . carried out algebraically? In Cartesian form this is [ r cos( ). up to a rigid motion? How do they determine its geometric parameters? Pedagogical Considerations: In general. r sin( )] . 2 y ) for each point ( x. yˆ ) ( x 2 y . How is such a rotation coordinates ( r . What sets of coefficients lead to the same curve. ) . we can take A cos . 1 5 / 4 A student who suspects that the curve is an ellipse should try to locate its vertices. y in terms of xˆ . approximately 2. yˆ y cos x sin .5) which might be recognized as precisely the eigenvalue problem for the matrix A (not T ). yˆ . y ) on the circle. Beginning with the circle x 2 y 2 r 2 . (1. B sin and simplify the equation of the parabola to ( x cos y sin )2 2C ( x cos y sin ) C 2 which immediately becomes yˆ 2 2Cxˆ C 2 . we will determine the rotation angle in a more geometric way and make the connection with eigenvalues. xˆ 2 2 xy 4 The question is whether this can be recognized as a rotated ellipse. One plausible way to do this is by using Lagrange multipliers to extremize xˆ 2 yˆ 2 subject to the equation as a constraint. the points of maximum and minimum distance from the center (which is at the origin due to the symmetry of the equation about the origin). y cos x sin ). and try to determine so that the equation simplifies. and this suffices to give the equation of the new curve as 5 ˆˆ yˆ 2 r 2 . a student investigates the effect of other simple (linear) transformations on the circle. 2 y ) for each point ( x. One could rotate this equation by an arbitrary angle . and what the rotation angle is. T 1 . What curve does she obtain? What equation could she have graphed to obtain the new curve directly? .5) with ( xˆ . 0 2 0 1 / 2 The quadratic form appearing in the equation of the ellipse is Xˆ t AXˆ .36 ( x cos y sin . This leads quickly to the equations xˆ yˆ xˆ xˆ (5 / 4) yˆ yˆ . where 1 1 A (T 1 ) t T 1 . Conversely. One way is to observe that the inverse rotation must be by angle . Note incidentally that the coordinate transformation in matrix notation appears as Xˆ TX . Students who have carried out the brute-force approach are quite likely to appreciate the benefit of referential symbolic reasoning. one could rotate the standard ellipse equation and compare with the example we have. y ) ( xˆ yˆ . Instead. The inverse of the coordinate transformation is ( x.117 . yˆ / 2) . yˆ ) ( x 2 y . Thus we need to re-express the equation of the parabola in terms of the new variables xˆ x cos y sin . where 1 2 1 1 T . she creates a new curve by plotting the new points ( xˆ . Another is to note that with A2 B 2 1 and tan B / A . The eigenvalues are (9 65) / 8 .133 and 0. yˆ ) and using the equation of the curve leads to Classroom Problem 2: Intrigued by the idea of obtaining an ellipse by squashing a circle. Referential symbolic reasoning can save a lot of work over the brute-force inversion of these equations to give x. Taking the dot product of the eigenvalue equations (1. students who are familar with diagonalizing symmetric matrices and their associated quadratic forms will have an easier time than those who are not. namely the major axis. and therefore the vertices can again be found from AXˆ Xˆ . so that the semiaxes should be r / . and would lead to a matrix having the same trace and determinant. is A C (the trace of M ).685 with its axes along the eigenvectors of A .750. they could locate the foci of the supposed ellipse along its axes and write down the condition for the sum of the focal distances to be the appropriate constant. Another method of arriving at the eigenvalue problem is to realize that the vertices of an ellipse are characterized as the only points on it where the normal vector is parallel to the position vector. To verify their conjecture. The position vectors of these points are eigenvectors of the symmetric matrix B / 2 A M . at this point students really must use the formulas for rotation of axes and explicitly transform the original equation into this simpler form. up to roundoff errors. Using the gradient of the quadratic form to find the normal vector.723 radians with the x -axis.685r . 2. namely 2. What sets of coefficients lead to the same curve. 0. However. The major axis would be along the unit vector (0. either as the points of maximum/minimum distance from the center (the origin. 1 2 . Those without this background may again start by locating the vertices of the (presumed) conic. 4 From this we learn that the sum of the eigenvalues. Once again. 12 . The type of conic depends on whether the . This would produce the equation of the curve once again. one obtains 2 AXˆ . We briefly discuss the simpler equation first.92 0. Students will now conjecture that the curve is a rotated version of the ellipse 2 2 x y 2 r .662) .92r and 0. which makes an angle of about 0. B /2 C The quadratic equation for the eigenvalues is B 2 4 AC 2 ( A C ) 0. the eigenvalues are always real. the Classroom Problem 3: What curve is represented by a general equation Ax 2 Bxy Cy 2 F 0 ? By a more general equation Ax 2 Bxy Cy 2 Dx Ey F 0 ? Describe the curves geometrically as completely as possible. up to a rigid motion? How do they determine its geometric parameters? product of the eigenvalues. Alternatively. by symmetry) or the points where the position vector is normal to the curve. The simpler equation 1 x 2 2 y 2 F 0 definitely represents a conic.37 xˆ 2 yˆ 2 r . is AC ( B / 2)2 (the determinant of M ). In particular. they could use the formulas for rotation of axes at this point. and the eigenvalues themselves are A C ( A C ) 2 B 2 / 2. Identify the type of conic section defined by the given equation (none of the conics are degenerate): a. 73x 2 72 xy 52 y 2 30 x 40 y 75 0 d. the linear terms cancel out if the linear equations 2 Aa Bb D. which is evident from its symmetry about the origin. it is a hyperbola when B 2 4 AC 0 and an ellipse when B 2 4 AC 0 . one eigenvalue vanishes and the corresponding semiaxis is undefined. if nondegenerate. . x 2 2 3 xy 3 y 2 2 3 x 2 y 0 b. Replacing x by x a . the plan succeeds unless B 2 4 AC 0 . It is still (generally) possible to remove the xy term by rotating one axis to lie along the eigenvector with the nonvanishing eigenvalue. In the remaining case B 2 4 AC 0 . and y by y b . Students should attempt to reduce the more general equation to the less general one by translation of coordinate axes to remove the linear terms. The semiaxes of the ellipse are | F / i | . In that case the locus is not symmetric about any origin. An equation of the given form cannot represent a parabola. The resulting locus. In fact. the original quadratic form is then a perfect square. and to remove the term linear in either x or y by a translation. so something is wrong. x 2 4 xy y 2 1 c. That is. and the “conic” is either empty or degenerate: a pair of lines. is a parabola. Ba 2Cb E are satisfied. Supplementary and Practice Problems: Problem 1.38 eigenvalues have the same or opposite sign. Use rotation of axes to graph the conic sections in parts (a)-(c). the plane z ax 1 leads to (m 2 a 2 ) x 2 m 2 y 2 2ax 1 0 . so the cone's equation must be z 2 m 2 ( x 2 y 2 ) . Fixing z in the equation of the cone must give the equation of this circle. The vertex of the cone is then at the origin and its axis is along the z -axis. Eliminating z between this and the equation of the cone gives a 2 m2 y x . The case may challenge students' visual intuition. How does the curve obtained depend on the location or inclination of the plane? This is an opportunity for students to develop their visual intuition for three-dimensional geometry and begin to connect it with algebraic representations. Eventually. visual intuition may suggest that the intersections are curves which are either closed and oval. and therefore it contains the two generators through these points and no point of any other generator. We have already seen that a horizontal plane cuts the cone in a circle (or a point if it passes through the vertex). as a rotating generator sweeps out the cone. One possible geometric explanation is as follows. when the cutting plane becomes parallel to a single generator of the cone. Visually. The various positions of the rotating line are called generators of the cone. The deformation continues until . which is therefore the cross-section of the cone by a horizontal plane at this height. In particular. the vertex angle is 2 . The circular intersection deforms into an oval shape and clearly remains closed. about the z -axis. so the interpretation is problematic. A representative example of such a plane is z ax with a m . or open and roughly parabolic. The xz plane meets it in the pair of lines z mx making angle 2 with each other. For example. for now they may continue intuitively. A point of the line at height z sweeps out a circle of radius | z / m | . m 0 . A direct algebraic approach is challenging. Other planes through the origin meet the cone in a single point if their inclination angle (the angle between the plane's normal and the z -axis) is less than . tan m . However. by continuity. For inclination angle we get a single generator of the cone. this is the equation for the projection of the intersection curve into the xy plane. z mx. The plane will meet a circular crosssection of the cone in two points. the intersection curve contains a single point on each generator. Therefore there are two moments when it lies in the plane. Imagine that an initially horizontal cutting plane slowly rotates to a position with . Considering planes not containing the vertex. The discussion can go in many directions. not its equation as a plane curve in the cutting plane. and any other plane containing the z -axis gives the same picture. and that the case m a does look parabolic. students may relate the coordinate x in the cutting plane to the spatial Cartesian coordinate x by x x cos . it passes from one side of the cutting plane to the other and back again. which it no longer meets: that point of intersection moved off to infinity as . m2 Together with z ax these are equations of two lines through the origin.39 Overarching supplemental problems Problem 1 Describe the possible intersections of a (double) cone with a plane. two generators of the cone. Its “elevation angle” is the inclination of the line. but for illustration let's adopt the definition that a (right circular) cone is the surface swept out by rotating a line through the origin in the xz plane. Students may observe that the relative size of m and a determines the sign of the x 2 term. This definition of a cone is thus equivalent to the union of lines joining the vertex to such a circle. find the locus of all points in the plane that are twice as far away from A as from B? How would you generalize this problem? Problem 5 Prove algebraically that. that every triangle has an inscribed circle (applicable to the ellipse case) as well as three excircles (in the other cases). The three medians (lines from a vertex to the midpoint of the opposite side) b. hyperbola). if the sphere is actually in contact with (“resting on”) the plane. due to Appolonius. the plane is parallel to two generators (the same two we saw when the plane contained the vertex) and also meets both nappes of the cone in open and unbounded curves. either by an intuitive argument in which one blows up a balloon enclosed between the cone and plane until it is tangent to both. Only the single nappe of the cone tangent to the sphere is relevant. The intersection curve is now open and unbounded. It may not be obvious at this stage that the position of the focus depends only on the conic section and not on the cone and plane used to obtain it. whether a circle determines its center. Problem 4 Given two point A and B in the plane. only one branch of a hyperbolic shadow appears. Explain why its shadow on a plane surface will be a conic section. hyperbola)? Given a point not on an ellipse (parabola. which is tangent to the sphere along a circle. a b . Every (nondegenerate) conic section arises as the shadow of a sphere in contact with the cutting plane. This will be the axis of a cone. How might we define the distance from a point to an ellipse (parabola. The three qualitatively distinct intersection curves are called ellipses (from ellipsis. or by the fact. and may necessitate a characterization of a conic section as a locus in its plane. Continuing to . we call the point of contact a focus of the conic section.40 the oval stretched. Now consider the sphere's shadow on a plane placed behind it. construct the point on the ellipse that realizes this distance. The elevation angle is given by cos R / D . and do not if internal.) Problem 2 A sphere is illuminated by a point light source some distance away. Rotate this line about the light source until it is tangent to the sphere. do not arise directly from the picture of slicing a cone. The three angle bisectors Problem 6 Show that the line connecting the topmost point to the rightmost point of the circle x2 y 2 given by x 2 y 2 a 2 b 2 is tangent to the ellipse 2 2 1 . Problem 3 We define the distance from a point to a line as the perpendicular distance. the following lines meet at a point. meaning deficiency or falling short). and hyperbolas (throwing beyond). As mentioned. It is now a generator of the cone. a. (The names. Light rays from the source reach this plane if they are external to the cone. extended if necessary) c. The three altitudes (lines through a vertex perpendicular to the opposite side. in any triangle. Rotating the generator about the axis produces the cone. where R is the radius of the sphere and D is the distance from the light source to the center of the sphere. parabolas (placing beside or comparable to). The boundary of the shadow is therefore the intersection of the cone with the plane. This is similar in spirit to Problem circle1. Draw the line through the light source and the center of the sphere. applied to a plane cross-section of the diagram. 41 Problem 7: Fix two points A and B in the plane. K. H. W. Johnstone and C. There follow a few which are less well-known. Computing the intersection of a plane and a natural quadric. What is the locus of all points P such that AP BP AB ? Students may recognize this as a degenerate case of the ellipse: the sum of distances is fixed by AB. Computers and Graphics 16(2)(1992)179 -. rather than merely the segment. Shene. Problem 8: A segment AB with endpoints on the sides of a right angle moves so that the distance AB remains fixed. they may incorrectly answer that the locus is a line. Macmillan 1875 .-K.186 . or take an unusual approach. . What is the locus of all the midpoints of AB ? B M A References There are innumerable sources for information on conic sections. J. This provides an opportunity to challenge their intuitive assumption that an equation derived from a locus is equivalent to that locus. Drew. A Geometrical Treatise on the Conic Sections. If students are not careful with the distances. Apostol and M. K. American Mathematical Monthly 115(9)(2008)795 -. Hansen.812 . . Shadows of the Circle. A. MAA 2005 . and directors. T.L. focal disks. M. World Scientific 1998 . New descriptions of conics via twisted cylinders. Kendig.42 V. Conics. Mnatsakanian.