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
h0
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.
5 1.1 The Circle
1.1.1 Focus Ways of Thinking and Ways of Understanding
 PGA way of thinking: Attending to interrelationships between physical/perceptual,
geometric, and algebraic realities.
 Algebraic invariance way of thinking. An equation can be rewritten in various forms
which make certain properties more noticeable. The link between the forms is
provided by something which remains invariant, for example the solution set.
 Algebraic way of thinking. This is a broad way of thinking. One of its instantiations is
the realization that when applying algebra, to a geometry problem for example, one
must “tell algebra” all the relevant geometric constraints. Likewise, we may talk of a
geometric way of thinking in this manner; namely, in solving a geometry problem,
one must “tell geometry” all the given conditions.
 Usefulness of completing the square.
 Understanding the concept of tangent line as a line intersecting a circle only once.
Understanding a property derived from this definition: the tangent line at a point is
perpendicular to the radius to that point.
We begin with an object familiar to students both perceptually and mathematically: the
circle. We draw a circle and ask, what makes this a circle? How do we recognize a circle or
communicate it to someone else? We hope to elicit the characterization as a locus: the set of
points in a plane equidistant from a specified point. This could be communicated by specifying
the center and radius (and the plane, if not obvious). This definition translates easily into the
usual Cartesian equation for a circle centered at the origin, or at any other point. The following
problems may be used in class to solidify these ideas through the necessity principle and the
repeated reasoning principle.
Classroom Problem 1: You drew a circle with a compass but forgot to mark its
center. It is possible to retrieve the center of the circle?
Classroom Problem 2: Some of the solutions to the previous problem led us to
conclude that an equation of the form ( x  a )2  ( y  b)2  r 2 represents a circle with
center ( a , b ) and radius r . Write down your meaning of the statement “a circle with
center ( a , b ) and radius r is represented by the equation ( x  a )2  ( y  b)2  r 2 .”
Explain again why this is the case.
Now consider the circle ( x  1)2  ( y  2)2  3 . Tami expanded this equation into the
equation x 2  y 2  2 x  4 y  2  0. Bruce didn’t see the original equation. Can Bruce
retrieve the center and radius of the circle from the second equation?
Classroom Problem 3: In the previous two problems, you learned that a circle with
center ( a , b ) and radius r has the equation ( x  a )2  ( y  b)2  r 2 .
The problem that Bruce encountered can be generalized: If we expand the brackets in
the circle’s equation and collect the corresponding terms, we can rewrite the above
equation in the form x 2  y 2  2ax  2by  ( a 2  b2  r 2 )  0 . This shows that the
equation of a circle can be written in two different forms: as
1. ( x  a )2  ( y  b)2  r 2
or as
6 2. x 2  y 2  2ax  2by  ( a 2  b2  r 2 )  0 .
We can make the second equation more compact and easier to remember by
substituting f  2a , g  2b , and h  a 2  b 2  r 2 to obtain
x 2  y 2  fx  gy  h  0. This shows that every circle can be represented by an
equation of the form
3. 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?
Classroom Problem 4: You are familiar with the concept “angle between two
intersecting lines.”
1. How would you define “angle between two intersecting planes”?
2. How would you define “angle between a line and plane” (if they intersect)?
3. How would you define “angle between two intersecting curves”?
4. We often talk about orthogonal lines, orthogonal planes, and a line
orthogonal to a plane. Can we talk about orthogonal curves?
Classroom Problem 5: Construct two orthogonal circles in one plane.
Classroom Problem 6: A circle of radius 4 is centered at (1, 2) . Where does it cross
the coordinate axes? What lines are tangent to it at these points?
Classroom Problem 7: Prove that the intersecting circles 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 ).
1.1.2 Pedagogical Considerations
A recurring theme of this unit (and, indeed, all the
units of this module) is that both geometry and
algebra are systems for drawing logical conclusions
from given data. To exploit the power of these
systems by drawing the strongest conclusions, it is
necessary to “tell geometry” or “tell algebra” all the
given conditions: the conditions must be stated in a form which these systems can process, and
all must be used nontrivially in the reasoning. This is what we referred to earlier as the algebraic
way of thinking and geometric way of thinking.
A geometric solution to this problem begins by choosing three points P1 , P2 , P3 on the
circle. Because three (noncollinear) points determine a circle, these three points suffice to “tell
geometry” which circle is given. To tell geometry that we want the circle through these three
points, rather than some other curve, we use the locus definition of the circle: the three points are
equidistant from the center. The locus of points equidistant from any two of the points is the
perpendicular bisector of the line segment joining them. Each of these bisectors therefore passes
through the center, which can be found as the intersection of any two of them. This solution is
obviously related to the theorem that the perpendicular bisectors of the sides of any triangle are
concurrent, in fact meeting at the center of the circumscribed circle. Traditionally, the center of a
circle is found by simply taking two non-parallel chords of the circle and finding the intersection
of their perpendicular bisectors. The two solutions are mathematically the same, of course.
However, the former solution is much more explicit about the reasoning process underlying the
Classroom Problem 1: You drew a
circle with a compass but forgot to
mark its center. It is possible to
retrieve the center of the circle?
7 construction—that there was a reason to think of the two chords. In our approach to this module,
we aim at demystifying mathematics for the students, in that solutions to mathematical problems
do not emerge from nothing but from careful reasoning and a representation of the necessary and
sufficient conditions of the problem.
There is an algebraic solution mirroring this geometric one. Let the point Pi have
coordinates ( xi , yi ) . The conditions telling algebra that the three points lie on a circle with center
( a , b ) and radius r are the three equations ( xi  a ) 2  ( yi  b) 2  r 2 . They state that each point is
r units from the center. Subtracting a pair of these equations gives a condition that the center is
equidistant from two of the given points, the specific distance r having cancelled out (referential
y  y2
x x 
x x 
  1 2 a  1 2  ,
symbolic reasoning). This turns out to be, for example, b  1
2
2 
y2  y1 
precisely the statement that ( a , b ) lies on the perpendicular bisector of the segment PP
1 2 . This is
the algebraic proof of the claim that the perpendicular bisector is the locus of equidistant points.
The center can be found by solving any two of these linear equations simultaneously. The third
linear equation is dependent on the two chosen, and therefore the center lies on it too.
Students may wish to locate the center as the intersection of two diameters. How easy this
is depends on the available tools. Given a ruler, one might locate two points on the circle a
maximal distance apart and join them. (Is this an allowed use of a ruler?) With straightedge and
compass, it can be done by inscribing a right angle at a point of the circle; its sides will meet the
circle at ends of a diameter. This could necessitate the theorem that an inscribed angle measures
half of the intercepted arc. It is useful for students to realize how geometric solutions depend on
the available tools.
In our classroom one student drew a circle on a sheet of paper and folded the paper so as
to bring one half of the circle onto the other, thus making the fold line a diameter. Two such
diameters intersect at the center. This is an excellent example of an empirical/perceptual proof as
opposed to a deductive one. The perceptual approach is valuable for highlighting intuitive
properties of the circle, which should be deductively verifiable later: its symmetry about any
diameter, for example.
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 , b ) and radius r .
Write down your meaning of the
statement “a circle with center ( a , b )
and radius r is represented by the
equation ( x  a )2  ( y  b)2  r 2 .”
Explain again why this is the case.
Now consider the circle
( x  1)2  ( y  2)2  3 .
Tami expanded this equation into the
equation x 2  y 2  2 x  4 y  2  0. Bruce
didn’t see the original equation. 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, and conversely, the ordered pair of
every point on the curve satisfies the equation.
Our experience suggests that this meaning is not
always clear to students. This issue goes beyond
the phenomenon that students are not careful to
distinguish a statement from its converse,
sometimes assuming that a chain of reasoning
proves an “if and only if” claim even though the
reasoning is not reversible. Having an equation
for some locus is so powerful that students are
unlikely to question the correspondence between
the locus and the equation—they see deriving an
8 Classroom Problem 3: In the previous two
problems, you learned that a circle with center
( a , b ) and radius r has the equation
( x  a ) 2  ( y  b) 2  r 2 .
The problem that Bruce encountered can be
generalized: If we expand the brackets in the
circle’s equation and collect the corresponding
terms, we can rewrite the above equation in the
form x 2  y 2  2ax  2by  (a 2  b2  r 2 )  0 .
This shows that the equation of a circle can be
written in two different forms: as
1. ( x  a )2  ( y  b)2  r 2
or as
2. x 2  y 2  2ax  2by  (a 2  b2  r 2 )  0 .
We can make the second equation more
compact and easier to remember by substituting
f  2a , g  2b , and h  a 2  b 2  r 2 to
obtain x 2  y 2  fx  gy  h  0. This shows
that every circle can be represented by an
equation of the form
3. 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, without separating the
steps of showing that (1) every point in the
locus satisfies the equation and (2) showing
the every point that satisfies the equation is
in the locus. To help students understand
this issue, we gave it repeated attention
when working with the different conic
sections.
Students worked on the second part
of the problem in groups and reconstructed
the original form of the circle's equation by
completing the square. This promotes the
Algebraic Invariance way of thinking and
introduces the usefulness of completing the
square, an essential technique throughout
this module (Repeated Reasoning).
Again students completed the square to
recast the equation as
f  
g
f 2  g2

x


y


 h . This

 

2 
2
4

represents a circle iff the constant on the
right side is a positive number r 2 , that is
iff f 2  g 2  4h . Students should ask (or
be asked) what is the locus if this condition
is not met: a single point, or else empty.
2
2
For the angle between two planes, students suggested
choosing a point in the intersection of the planes,
choosing a line through this point in each of the
planes, and taking the angle between these lines.
They realized that this concept is not well-defined,
because it depends on the choices made. They
decided instead to use the angle between the planes’
normal lines, which is not ambiguous. This
productive disequilibrium is a very important
experience in terms of the origin of mathematical
definitions. First, new concepts often rely on old
ones: the angle between planes is defined in terms of
the already-understood angle between lines.
However, if a definition allows different people to
make different choices, which lead to different
results, then the Need for Communication requires
that these choices be standardized in some way so that everyone obtains the same answer.
Similarly, the angle between a line and a plane cannot be defined as simply the angle
Classroom Problem 4: You are
familiar with the concept “angle
between two intersecting lines.”
1. How would you define “angle
between two intersecting planes”?
2. How would you define “angle
between a line and plane” (if they
intersect)?
3. How would you define “angle
between two intersecting curves”?
We often talk about orthogonal lines,
orthogonal planes, and a line
orthogonal to a plane. Can we talk
about orthogonal curves?
9 between the line and any chosen line in the plane which meets it. The angle between the line and
its orthogonal projection into the plane will do, and this is equivalent to taking the complement
of the angle between the line and the plane’s normal.
To define the angle between curves, and specifically the idea of orthogonal curves,
students suggested using their tangent lines. Asked why this makes sense, they responded that
the lines locally approximate the curves and we already know what the angle between lines
means. This is a proper answer. The definition for the angle between two curves emerged
naturally for the students from their image of tangent line as a local approximation to a curve.
Here the instructor can discuss the relationship between this meaning of “tangent line” and
Euclid’s meaning as a line intersecting exactly once (which may be familiar to students from
high-school). This is part of the need for communication, in that different meanings of the same
term must be equivalent, for otherwise it wouldn’t be possible to communicate meaningfully
about the term. Furthermore, this would advance another important way of thinking—that a term
can have multiple interpretations, and it is advantageous to have multiple interpretations for a
term. Once this equivalency has been established, the instructor may ask, “Imagine that I am
blind. How can I prove that given a circle, there is a line that does not intersect it?” The “game”
of playing blind is to remove the use of visual perception, which is common among students. A
geometric approach to this depends on a careful axiomatization of geometry, so algebra is more
appropriate here. If the circle and line have equations ( x  a )2  ( y  b)2  r 2 and y  mx  k ,
how many intersection points can they have? Students should explain clearly why substituting
y  mx  k into the equation of the circle has the meaning of locating intersection points. Since a
quadratic equation results, there are two, one, or no intersection points, and any of these
possibilities can occur by adjusting the parameters. The case with no intersections answers the
given question, and the case of a single intersection defines the idea of a tangent line.
In our classroom, a student presented her approach
(see Figure 1). Circles centered at O and O 
intersect at A and B ; lines OO  and AB are
drawn and meet at C . She suggested that the radii
must be greater than half OO  in order for the
circles to intersect. Such conjectures should always
A
be examined carefully, using perceptual evidence,
relevant theorems, potential counterexamples, and
so forth, so that students progressively develop
O
C
O'
their skepticism and standards of proof. 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. The student made several further
conjectures, 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), OO   AB and bisects AB ; the circles will be orthogonal if
their radii are equal.
A second student presented this reasoning. Calling the radii r1 , r2 and the distance
Classroom Problem 5: Construct
two orthogonal circles in one plane.
10 between the centers s , the circles will be orthogonal if each radius to A or B is tangent to the
other circle, that is, if the radii are perpendicular there. This happens if r12  r22  s 2 (by the
converse to the Pythagorean theorem). Thus, the construction can be made as follows. Draw one
circle with any center O and radius r1 ; at the endpoint A of a radius, draw a perpendicular
segment AO  of any length r2 , and a second circle centered at O  will be orthogonal to the first
circle at A . Why also at the second intersection B ? By the earlier argument that there are pairs
of congruent triangles! So students see that the first approach was not “wrong” or a waste of
time; the reasoning is valuable after we see how it fits into the structure of the problem.
The equation of the circle is ( x  1)2  ( y  2)2  16 ,
from which the intercepts are easily found as
(0, 2  15) and (1  2 3,0) . They also follow from
applying the Pythagorean theorem to a diagram,
which should not be surprising since the Pythagorean
theorem is the basis for the equation of the circle.
They are of course symmetric about the vertical and horizontal diameters of the circle.
To find the tangent lines, we need to tell algebra or geometry what a tangent line is. As
we saw, for circles one can use the simplest definition: a line meeting the circle in exactly one
point. However, students should have no trouble giving examples of lines which meet a parabola
in one point but are not tangent to it, or lines tangent to a non-convex closed curve which meet it
more than once.
As Euclid shows, the line perpendicular to the radius of a circle at some point P on the
circle meets the circle only there and is otherwise external to it. Indeed, if O is the center and Q
some other point on this line, then OQ is greater than the radius OP by the Pythagorean
theorem, so Q is outside the circle. This can be confirmed by students who know calculus:
implicit differentiation of the equation of the circle gives dy / dx  ( x  1) / ( y  2) , the negative
Classroom Problem 6: A circle of
radius 4 is centered at (1, 2) . Where
does it cross the coordinate axes?
What lines are tangent to it at those
points?
reciprocal of the slope of the radius to ( x , y ) . The slopes are 1 / 15 at the y -intercepts and
 3 at the x -intercepts, which look plausible from a picture. The two signs again reflect the
symmetry of the circle about its diameters. Since the equation of the circle presents it as a level
set of a function f ( x, y ) , one can also find a normal vector by taking the gradient, which gives
twice the radius vector. One can finesse the meaning of the tangent line by simply defining it to
be the line having the slope computed by calculus, but this gives no insight unless the process of
differentiation is understood.
An algebraic approach may lead to more insight than calculus for many students. We
seek a line y  mx  b intersecting the circle only once, say at a y-intercept x  0 . 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. If this has only one root x  0 , then two
coefficients must vanish: b 2  4b  11  0 gives b  2  15 , so that the line and the circle have
the same y-intercept, and then m  1 / (b  2)  1 / 15 , the expected slope.
11 Orthogonality condition for circles. This is a beautiful
example of the PGA WoT (way of thinking): a natural
geometric condition, orthogonality of circles, is equivalent
to a simple algebraic condition on the coefficients in their
x 2  y 2  f 2 x  g 2 y  h2  0 are
equations. 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 ).
two distinct implications to prove, an idea that will recur
(Repeated Reasoning!) often. If the circles are orthogonal,
with radii r1 , r2 and centers separated by s , then the Pythagorean Theorem implies r12  r22  s 2 .
Completing the squares in the equations of the circles, we find their radii given by
1
1
r12  ( f12  g12 )  h1 , r22  ( f 22  g 22 )  h2 . Their centers are ( f1 / 2, g1 / 2) and ( f 2 / 2, g 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 ). Substituting indeed
4
produces the claimed algebraic condition. Conversely, if the condition holds we can reverse the
reasoning and deduce r12  r22  s 2 , whereupon the converse to the Pythagorean Theorem shows
that the circles are orthogonal. Students may not realize that the converse to the Pythagorean
Theorem is distinct from the theorem itself and requires a distinct proof. It is useful to keep track
of the results from geometry that are being used to support the reasoning throughout this module,
and students should independently make sure they can prove them. Of course, there are calculusbased approaches to this problem using the slopes of the tangent (or normal) lines to verify
orthogonality.
A final question for students, to clarify the subtle logic in this problem: Does the given
algebraic condition ensure that the circles intersect, or does it only make them orthogonal
assuming they intersect? Since the condition implies r12  r22  s 2 , if follows that there is a right
triangle having sides r1 , r2 , and s . Then the triangle inequality says r1  r2  s and | r1  r2 |  s ,
which means that the centers of the circles are close enough for them to intersect. So the
condition does guarantee that the circles intersect, and orthogonally. However, it does not
guarantee that the “squared radii” computed from the equations are actually positive; this has to
be assumed.
Classroom Problem 7: Prove
that the intersecting circles
x 2  y 2  f1 x  g1 y  h1  0 and
1.1.3 Supplementary and Practice Problems
The following problems should be seen as a continuation of the Classroom Problems.
Problem 1. Choose any values for f , g , and h in the equation
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 . Now choose another set of values for f , g , and h , but
this time f should be a fraction, g an irrational number, and h a negative integer.
Again convert the resulting equation into an equation of the form
( x  a ) 2  ( y  b) 2  r 2 .
Problem 2. Find the equation of a circle centered at the origin that is tangent to the
line 2 x  2 y  39 .
Problem 3: Let P, Q, R, and S be four distinct points in the plane such that no three
12 of them are collinear.
a. How many circles can you draw through the point P?
b. How many circles can you draw through the points P and Q?
c. How many circles can you draw through the points P, Q and R?
d. How many circles can you draw through the points P, Q, R and S?
Problem 4: Find the intersection points of the circles ( x  1) 2  ( y  1) 2  5 and
x 2  ( y  1) 2  4 . 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.
Problem 6: How many common tangents can there be for two circles in the plane?
Exhibit all the possibilities, and prove that there are no others.
Problem 7: Given 2 intersecting circles in a plane, consider the 4 tangents at the
points of intersection (1 tangent at each of 2 points for each of 2 circles). Draw
perpendicular lines from each circle center to each tangent, thus obtaining 6 points of
intersection: 4 lines from each center yields 8 points, only 6 of which are distinct.
Show that all these points (A,B,C,D,E,F) are on the same circle.
Problem 8: We have seen that an equation of the form ( x  a ) 2  ( y  b) 2  r 2
represents a circle in the plane with center ( a , b ) and radius r .
a. Define sphere geometrically.
b. What is the equation of a sphere with center ( a, b, c ) and radius r ?
c. Given two equations of two spheres, find an algebraic condition which
ensures the two spheres (i) do not intersect, (ii) intersect at exactly one point,
(iii) intersect at more than one point.
d. For the last case, where the two spheres intersect at more than one point, what
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. Some of the
following problems serve to do so. In Problem 8, for example, students need to use the definition
of a circle, including the part about being planar, to verify that the given curves are circles. They
also need to observe that parametric equations describe not only a locus, but also the motion of a
point along that locus in time. This motion is useful for physical modeling but irrelevant for
geometric properties of the locus. The meaning of the parameter t needs to be deeply understood
in order to distinguish between intersection of the loci and collision of the points moving along
them. Here algebra encodes more information than the geometry requires.
Problem 10: A particle moves along the line 2 x 
3
y  2  0 . At 12:00 AM, the
5
2
. 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. What is the
9
particle’s position (in the x - 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 ), y (t ), z (t )) is often written as
r (t )  x(t )i  y (t ) j  z (t )k . Consider the two parametric curves:
r1 (t )  (cos t )i  (sin t ) j and r2 (t )  (cos t ) j  (sin t )k . What are these curves? Do they
intersect?
Problem 12: A particle moves so that its position in the plane at time t seconds from
the start of its movement is (cos t ,sin t ) . If the particle’s movement was observed for
2 seconds, what curve did the particle complete? If the particle’s movement was
particle’s x -coordinate is

seconds, what curve did the particle complete? Answer the same
2
question for 4 seconds and for 1 second.
Problem 13: Three particles, A, B, and C, move in space. The positions of the three
objects in space at time t seconds from the start of their movement are, respectively:
t
(cos t ,sin 2t , 0) , (cos , 0,sin t ) , (0,cos 2t ,sin 4t ) . 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. Where are they
located t minutes after noon? What is the angle between them at that time, and what
is the distance between their tips? At what times do the hands coincide?
observed just for
14 2. Unit 2: The Ellipse
Focus Ways of Thinking and Ways of Understanding







Referential symbolic way of thinking. Attending, when there is a need, to the meaning
of symbols and their manipulations.
Deductive reasoning: Logical structure of proofs: what is given, what is proved, what
can be assumed or chosen freely. Distinguishing between a theorem and its converse.
Algebra as deduction: Every algebraic computation is a proof of something. Try to
state explicitly what it proved.
PGA way of thinking: Attending to interrelationships between physical/perceptual,
geometric, and algebraic realities.
Algebraic invariance way of thinking. An equation can be rewritten in various forms
which make certain properties more noticeable. The link between the forms is
provided by something which remains invariant, for example the solution set.
Algebraic way of thinking. This is a broad way of thinking. One of its instantiations is
the realization that when applying algebra, to a geometry problem for example, one
must “tell algebra” all the relevant geometric constraints. Likewise, we may talk of a
geometric way of thinking in this manner; namely, in solving a geometry problem,
one must “tell geometry” all the given conditions.
Usefulness of completing the square.
Classroom Problem 1: What is an ellipse? [See pedagogical discussion of this problem.]
Classroom Problem 2: Show that the major axis (the chord containing the foci) of an ellipse is
greater than the minor axis (the chord through the center perpendicular to the major
axis).
Classroom Problem 3: Show that any chord AB of an ellipse passing through the center is
bisected by the center.
Classroom Problem 4: Find a Cartesian equation of an ellipse. Does it confirm the intuition
that an ellipse can be obtained by “squashing” a circle?
Classroom Problem 5: We know that the tangent line to a circle is perpendicular to the radius
at the point of tangency. Is there a similar characterization of the tangent line to an
ellipse?
Classroom Problem 6: Is the focus/directrix definition of an ellipse equivalent to the two-focus
definition? That is, is the locus of points whose distances from focus and directrix are
in the ratio 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,
formalize a geometric definition capturing it, and represent it
algebraically in a Cartesian coordinate system. We asked students: what other geometric objects
might we study in this way? One student suggested an ellipse, for which the intuition is a kind of
oval. Another student suggested a formal definition, which the instructor rephrased: Fix two
points F1 , F2 in a plane, and a number d  F1F2 . An ellipse is the locus of points P in the plane
Classroom Problem 1:
What is an ellipse?
15 such that PF1  PF2  d . The instructor drew one on the board using a shoelace wrapped around
two pegs (perceptual verification that the definition is compatible with our intuition).
This problem serves to check that our formal definition
can justify properties of the ellipse that we intuitively
expect to hold. Let T , T  be the vertices on the major
axis with T closer to F2 , A a vertex on the minor axis,
Classroom Problem 2: Show that
the major axis of an ellipse is
greater than the minor axis.
and O the center (the midpoint of F1F2 ). Define
a  AF1  AF2 so that d  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.
T
c  OF1  OF2 . Then AF1  AF2  2a, also
TF1  TF2  2a, and a  b . The conclusion would
follow from TO  a , which would follow from
TF2  T F1 . This follows from TF1  TF2  T F1  T F2
upon substituting TF2  2c for TF1 and T F1  2c for
T F2 .
Another intuition about the ellipse to verify is that any
chord AB through the center O should be bisected by
the center (symmetry). To prove AO  OB , how do we
tell geometry what is given? That A and B are on the
ellipse is expressed by AF1  AF2  BF1  BF2 . Also,
from the definition of the center, OF1  OF2 . Students
A
conjectured that AF1  BF2 but could not justify it; also
F1
that F2 AF1 is bisected by AB , which is not true.
F2
However, AOF1  BOF2 since they are vertical
angles. After several unsuccessful attempts by students,
A'
the instructor suggested a proof by contradiction: what
B
disaster follows if, contrary to expectation, AO  OB ?
Then choose A on OB with OA  OA (see Figure 3).
Figure 3: Bisected ellipse chord From congruent triangles it now follows that both B
and A are on the ellipse; that is, AF1  AF2  BF1  BF2 . This is contrary to intuition, because
A is inside F1BF2 , so the sum of its distances to the vertices F1 and F2 must be less than B 's.
The intuition is confirmed by Euclid's Proposition I.21, whose proof the instructor reviewed.
The instructor asked students: what is the pedagogical value of this proof? Students suggested:
present facts as they are needed in the proof; extract subdiagrams from diagrams and draw them
separately; the need for indirect proof was unexpected; the steps must still be assembled into a
complete and organized argument [do so!]. The instructor added: sometimes the teacher is an
actor, sometimes he is genuinely thinking at the board; the teacher is human and can demonstrate
getting stuck and yet recovering. In this case, the instructor knew the status of student
conjectures but did not reveal this fact. Part of the important lesson from this problem was that
proofs are not done just to have a proof, but also to learn new things along the way.
O
16 Closely related to this problem is the claim that the ellipse is symmetric about its center.
We can prove this, and in fact more: the ellipse is symmetric about both axes. Reflection of any
point P on the ellipse about the line joining the foci (the major axis) preserves both focal
distances and so maps P to another point on the ellipse. Reflection about the perpendicular
bisector of the focal segment (minor axis) interchanges the two focal distances and so preserves
their sum. The composition of these reflections is reflection about the center, so the ellipse is
symmetric about its center.
Classroom Problem 4: Find a
Cartesian equation of an ellipse.
Does it confirm the intuition that
an ellipse can be obtained by
“squashing” a circle?
For this problem, we translate our definition of ellipse
into algebra. We will later consider general position and
orientation of the ellipse, but for now putting the origin
at the center and the foci at (  c, 0) , the definition gives
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a. (1.1)
As c  0 this becomes the circle x 2  y 2  a 2 ;
visually, the ellipse seems to be a squashed version of this circle. Indeed, squashing the y
coordinate by a factor b / a should produce the correct minor axis, and gives the new equation
( x 2 / a 2 )  ( y 2 / b2 )  1 , which a student had remembered. The class therefore conjectured that
this equation is equivalent to (1.1). Now there is a Need for Computation to verify the conjecture
by reducing (1.1) to this simpler form. The algebra is lengthy but straightforward and uses
a 2  b2  c 2 .
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 ,
(1.2)
r22  ( c  x ) 2  y 2 .
Subtracting gives the simple result r12  r22  4cx , which is productive because we recognize the
difference of two squares. Factoring and recalling r1  r2  2a gives r1  r2  2cx / a . Knowing
the sum and difference of the focal radii, we can solve for each, obtaining the unexpectedly
simple formulas
cx
r1  a  ,
a
(1.3)
cx
r2  a  .
a
Substituting back in (1.2) produces the standard form of the equation of the ellipse. This
reasoning is simpler and more insightful than the previous computation. Allowing students to see
this promotes the Referential Symbolic WoT, which we continued to promote throughout the
module. Students should not fear lengthy algebra, but should pause often to unpack the meaning
of the symbols and equations, and be alert to the possibility of meaning-based simplification.
The squashed circle viewpoint is very useful for PGA reasoning. For example, we
showed that a circle determines its center; does an ellipse determine its foci? The ellipse is
obtained by squashing a circle whose diameter is the major axis, in the direction perpendicular to
this axis. It follows that the major axis is uniquely determined as the longest chord through the
ellipse, and the minor axis as its perpendicular bisector. Their lengths, 2a and 2b , determine c
and hence the positions of the foci. (See Supplementary Problem 1 for a Euclidean construction
17 of the foci.)
In our classroom students briefly considered using the
radius from the center, but preferred the focal radii.
They conjectured that the angles between the tangent
line and these may sum to 90 , but then switched to the
conjecture that these angles are equal. Students
observed that this is true in the special case of a circle,
and at the vertices of an ellipse, providing some initial
evidence for the conjecture. Let a line be tangent to an ellipse with equation in standard form at
point P  ( x1 , y1 ) in the first quadrant, and make angle  with the major axis (see Figure 4). Let
Classroom Problem 5: We know
that the tangent line to a circle is
perpendicular to the radius at the
point of tangency. Is there a
similar characterization of the
tangent line to an ellipse?
the angles between the tangent line and the focal radii PFi be  i . Let the angles between
PF1, PF2 and the major axis be  ,  respectively. 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, each promoting important Ways of Thinking,
and the different approaches can be productively compared. First, we need to determine the
equation of the tangent line, that is, tell algebra that the line is tangent to the ellipse. This can be
done using calculus, or the algebraic method of finding a line having a single intersection of
multiplicity two with the ellipse. The result is ( x1 x / a 2 )  ( y1 y / b 2 )  1 . The notational
distinction between ( x1 , y1 ) , the chosen point of tangency, and ( x , y ) , a variable point on the
tangent line, deserves emphasis. The Need for Communication requires distinguishing them;
using the same symbol for two objects would tell algebra that they are the same. ( x1 , y1 ) satisfies
both the equation of the ellipse and that of the tangent line; ( x , y ) satisfies only the latter. Some
angles in Figure 4 (more precisely, their tangents) can be computed from the slopes of the lines:
tan   b 2 x1 / a 2 y1 ,
tan   y1 / ( x1  c),
tan   y1 / (c  x1 ).
18 We can use the exterior angle theorem to find  1 and  2 :  1     and  2     . Then trig
identities [prove them!] allow us to compute, for example,
tan   tan 
tan  1 
.
1  tan  tan 
Here is an example of the Necessity for such trig identities (“When will I ever use this?”).
One finds straightforwardly (don't fear complex algebra!)
b 2 x1
y
 1
2
a y1 x1  c
tan  1 
,
b 2 x1
1 2
a ( x1  c)
y1
b2 x
 2 1
c  x1 a y1
tan  2 
.
b 2 x1
1 2
a (c  x1 )
We first clear denominators to obtain
b2 x1 ( x1  c )  a 2 y12
tan  1 
,
y1[a 2 ( x1  c )  b2 x1 ]
a 2 y12  b2 x1 ( c  x1 )
.
y1[a 2 ( c  x1 )  b2 x1 ]
One can check the equality of these by brute force (e.g. multiplying one by the reciprocal of the
other and simplifying), but the Referential Symbolic WoT suggests a meaning-based
simplification. Indeed, from the equation of the ellipse, the terms b 2 x12  a 2 y12 in the numerators
tan  2 
simplify to a 2b 2 , and in the denominators a 2  b 2  c 2 can be used. Each tangent then reduces to
b2
, 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, it should have no units and thus
the numerator and denominator must have the same overall power). Finally, we need to consider
whether tan  1  tan  2 suffices to conclude that  1   2 . This provides Necessity for the concept
of a one-to-one function on a specific domain, here 0     .
A second proof of the equal-angles property of the tangent line requires less computation
and provides more insight. It suffices to show that the normal line at the point P of the ellipse
bisects angle P in the triangle F1PF2 . If this normal meets the opposite side F1 F2 at a point K ,
then by Euclid, Prop. VI.3, this is equivalent to showing that F1K / F2 K  F1P / F2 P . (An angle
bisector in a triangle is characterized by dividing the opposite side in proportion to the adjacent
sides.) Since the normal line has equation [Prove this!] ( xa 2 / x1 )  ( yb 2 / y1 )  c 2 , the point K
has coordinates ( c 2 x1 / a 2 ,0) and the required proportion becomes
19 c  ( c 2 x1 / a 2 ) r1
 ,
c  ( c 2 x1 / a 2 ) r2
which follows from our earlier computation of the focal radii, (1.3). There is a third, physicsbased proof which provides insight. Imagine an object moving around the ellipse, with any
velocity v , not necessarily constant. Since the sum of its distances to the foci remains constant,
the sum of the rates of change of the focal distances is zero. But the rate of change of the
distance to a focus is simply (minus) the component of velocity in the direction of the focus. This
says that v cos  1  v cos  2  0 , again implying  1   2 .
Finally, a useful follow-up question is: given an ellipse with its foci, perform a Euclidean
construction of the tangent line at a specified point. Besides the solution following from the
equal-angles property, namely drawing the focal radii and bisecting the angle they form, there is
another following from the squashed circle viewpoint. Begin with the circle x 2  y 2  a 2 and its
tangent line at some point ( x1 , y1 ) . 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 , by1 / a ) , and the squashing does not change the x -intercept of the tangent line. Therefore,
from the tangent line to the circle we obtain this intercept, and the tangent line to the ellipse is
drawn by joining the intercept to P .
Classroom Problem 6: Is the
focus/directrix definition of an
ellipse equivalent to the two-focus
definition? That is, is the locus of
points whose distances from focus
and directrix are in the ratio e  1
The results obtained earlier for the focal radii, e.g.
r2  a  ( cx / a ) , are surprisingly simple--simple enough
that the PGA WoT suggests seeking a geometric
interpretation. A linear function of x can be interpreted
in terms of the distance from ( x , y ) to a vertical line. In
this case, r2  e( f  x ) , where we define e  c / a and
f  a 2 / c . That is, r2 is e times the distance from ( x , y ) to the vertical line l with equation
x  f . [We observe that f  a , so that this line lies to the right of every point on the ellipse,
and also that e  1 .]
A useful Way of Thinking is that every algebraic computation is a proof of something,
and one should try to state explicitly what it proved. Our computation of r2 proved that, if P lies
on the ellipse, then PF2  ePl . That is, the distance from P to the focus F2 is e (the
eccentricity) times the distance to l , the directrix.
P
Most students do not yet possess the WoT that the
converse should automatically be investigated, and
that reversing the algebraic reasoning may prove it.
r1
r2
Here we ask whether a point satisfying PF2  ePl
is necessarily on the ellipse. If so, then we have an
c
x
O
F1
F2
alternative definition of an ellipse, the
focus/directrix definition. 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
20 r12  r22  ( c  x ) 2  ( x  c ) 2  ( a  ac x ) 2 . Since a  ac x  0 for every point on the ellipse [prove
it!], we can take the square root to obtain r1  a  ac x , and add to obtain r1  r2  2a . Thus, the
focus/directrix property implies the two-focus definition of the ellipse.
In 1822, G.P. Dandelin discovered a beautiful proof
that “conic sections” are indeed plane sections of a
cone. In our classroom we motivated this
construction and presented the reasoning in the case of the ellipse. It is also possible, but
complicated, to take an algebraic approach: obtain equations for a cone and a plane cutting it in
three-dimensional Cartesian coordinates, and derive an equation for their intersection. In addition
to the complexity of the algebra, a conceptual difficulty is that the three-dimensional Cartesian
coordinates x, y are not also standard Cartesian coordinates in the cutting plane, where we want
to have an equation for the locus.
Intuition tells us that a “horizontal” plane cuts a cone in a circle, whereas if the plane is
tilted slightly the intersection is an oval curve which may be an ellipse. Perceptually, it seems
unlikely that this oval will be symmetric about its center rather than egg-shaped; we need
deductive reasoning to correct our intuition in this case. In Figure 6, a cone has vertex V and is
tangent to a sphere along circle C lying
in plane  and having center O . A
plane  meets the cone in the curve E
(which we assume is closed), is tangent
to the sphere at F , and cuts the plane
 in the line d . Conversely, for any
plane  cutting the cone in a closed
curve E , we can draw a sphere tangent
to both the cone and the plane as shown.
Our objective is to show that E is an
ellipse. P is an arbitrary point on E ,
the generator VP meets  at L , and
PM is drawn perpendicular to  . PD
is drawn perpendicular to d . The
cutting angle  is PDM and the
cone's elevation angle  is VLM . It
is important for students to attend to the
temporal (logical) sequence in which
elements are added to the diagram:
which points, lines, and so forth are
chosen first and which others are then
determined by these choices. Some
elements are given, such as the cone and
plane, some are fixed, like the point of
Figure 6: Dandelin sphere construction of ellipse
tangency F , and some are variable like
P . We introduce the variable point P in order to show that however it is chosen on E it will
satisfy one of the definitions of an ellipse. Students need practice (Repeated Reasoning) with the
overall logical structure of such proofs.
Classroom Problem 7: Is a (suitable)
plane section of a cone an ellipse?
21 To prove that E is an ellipse, we need to identify candidates for the focus and directrix.
The point of tangency F might be the focus, and d might be the directrix. Some support for this
conjecture comes from the facts that, as the plane becomes horizontal, F moves to the center of
the circle E and d moves off to infinity. By way of telling geometry that the sphere is inscribed
in the cone, we note that PL  PF are tangents to the sphere from the same external point.
(Students are familiar with the fact that tangents to a circle from an external point are equal, but
had difficulty deducing the corresponding fact for spheres. To obtain the latter from the former,
one has to find the right circular cross-section of the sphere.) We also have
PM / PL  PM / PF  sin  and PM / PD  sin  . Therefore,
PF sin 

 e,
PD sin 
a constant independent of P . If we know that e  1 , then we have shown that E is an ellipse.
This follows from our implicit assumption that the plane cuts the cone in a closed curve, which
occurs only when the plane cuts each generator of the cone. That in turn requires    , which
implies e  1 .
Although Dandelin's proof is visual and elegant, it also raises several subtle issues which
should be made explicit if not necessarily resolved rigorously in class. First, how do we know
that there is always a (unique) sphere tangent to both the cone and the plane  ? An intuitive
continuity argument is convincing but not rigorous: begin with a small spherical balloon tangent
to the cone only, and ``blow it up" until it becomes tangent to the plane as well. A geometric
proof can be given by using a suitable plane cross-section of the three-dimensional diagram and
the fact that a triangle has a unique inscribed circle. Second, the fact that a cutting plane with
   produces a closed curve of intersection relies on some intuitive topology. (Indeed, the
proof shows only that the curve of intersection is contained in an ellipse. We are relying on this
intuitive topology for the conclusion that the curve is a complete ellipse.) We did not press the
issue of what it means to be a closed curve at this point, but we returned to it when comparing
the Dandelin sphere proofs for the different conic sections. Third, the definition of a cone as a
figure of solid geometry may still be at the intuitive level for students and should be formalized.
For example, begin with a circle and a point other than its center on the line through its center
and perpendicular to its plane; take the union of all lines joining this point to the points of the
circle. Students do not always realize initially that this defines a double cone.
Supplementary and Practice Problems:
Problem 1. Given an ellipse, construct (synthetically) its foci.
Although the construction is purely synthetic, the proof uses coordinates. First, we find
the center. Construct a pair of parallel chords (AB and CD in Figure 6). Connect the midpoints
M 1 and M 2 . Construct a second pair of parallel chords EF and GH not parallel to the first pair,
and connect their midpoints M 3 and M 4 . Then the intersection of M 1M 2 and M 3 M 4 is the
center of the ellipse. To prove this, it suffices to show that the center is contained on the line
joining midpoints of any 2 parallel chords. Let the ellipse be given by ( x / a )2  ( y / b)2  1 .
Suppose one chord is given by the line y  mx  n1 and the parallel chord is given by y  mx  n2
. 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 . However, we can find the midpoint without
22  mn1a 2
(average of 2 solutions to a
b2  m 2a 2
 m 2n1a 2
 n1 . Similarly, the other midpoint will
quadratic is  B / 2 A) , so y-coordinate is y1  2
b  m 2a 2
  mn2a 2 m 2 n2 a 2

be ( x2 , y2 )   2
, 2
 n2  . 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, 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
,

ma 2
F
which is independent of ni , giving the desired
M1
equality.
To find the axes, we pick any point on
O
the ellipse and draw a circle centered at O
M2
through that point. The circle will intersect the
G
ellipse at 4 points, which we connect to get a
M3
rectangle whose sides are parallel to the
B
H
desired axes [prove this!]. We can then draw
D
lines parallel to these sides through the origin.
Figure 7: Constructing ellipse foci 1
With the axes in hand, pick a point P on the
ellipse that is also on the minor axis. Draw a circle centered at P with radius equal to the semimajor axis. The intersection points of this circle with the major axis are the foci.
The most difficult step in this proof can be viewed as showing that the locus of midpoints
of a family of (all possible) parallel chords of an ellipse is a diameter of the ellipse. Analogous
statements for the parabola and hyperbola can be proved similarly.
Problem 2. Use a second Dandelin sphere to confirm that an ellipse has a second
focus/directrix pair.
Problem 3. Let E be an ellipse with center O , major axis of length 2a , and minor axis of
length 2b . Prove that if P  E , then b  PO  a .
Problem 4. Find the equation of the ellipse with foci   23 ,0  and  23 ,0  and directrix the line
x  4.
Problem 5. Find the eccentricity and the directrices of the ellipse
2
2
 x  2   y  3


 
  1.
 5   4 
Problem 6. Consider the equations
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.
a. Show that one of the above equations defines an ellipse and that the other equation does
not define an ellipse.
b. For the equation which does define an ellipse, find the coordinates of the center, the
coordinates of the foci, the length of the major axis, and the length of the minor axis.
y2
x2
Problem 7. Find an equation for an ellipse with the same foci as the ellipse 25
 16
 1 but
greater eccentricity. Graph both ellipses on the same set of axes.
Problem 8. Let E be an ellipse with center O . Let P  E . Find an expression for PO in terms
of the angle between PO and the major axis of E .
Problem 9. Find a tangent line to the ellipse x 2  4 y 2  196 which is perpendicular to the line
x  5.
Problem 10. Find a normal line to the ellipse 3x 2  2 y 2  50 which is perpendicular to the line
6x  y  7 .
Problem 11. Prove that if PQ is a diameter of an ellipse, then the tangent lines to the ellipse at
the points P and Q are parallel.
Problem 12. Let PQ be a diameter of an ellipse.
a. Prove that the midpoints of the chords of the ellipse which are parallel to PQ all lie on
the same line.
b. Prove that the line containing the midpoints of the chords of the ellipse which are
parallel to PQ is parallel to the tangent lines to the ellipse at the points P and Q .
Problem 13. Is a plane section of a circular cylinder also an ellipse?
Perceptually, it seems unlikely that sections of cylinders and cones have the same shape.
The Dandelin sphere argument actually applies to the cylinder as well as the cone. Again, this is
an excellent opportunity for students to appreciate the advantages of the Deductive over the
Empirical proof scheme.
Problem 14. A mining company drills a mineshaft into the ground. The hole is a circular
cylinder, with its axis making an angle  with the vertical. If the (elliptical) opening at the
surface must fit within the square plot of ground owned by the company, whose side is twice the
diameter of the cylinder, how does this restrict  ?
If the axes of the ellipse are assumed parallel to the sides of the square, this is an easy
application of the squashing of the circle to an ellipse. However, a larger ellipse will fit
diagonally. The sides of the square are then tangent to the ellipse and make a 45 degree angle
with the axes. Determining where such lines are tangent is a nontrivial exercise.
If we tilt our heads by the angle  , we see the cylindrical mineshaft as if its axis were vertical,
and the ground as a cutting plane with elevation angle  . Then the elliptical opening can be
described by the equation x 2 cos2   y 2  r 2 , where r is the radius of the hole. Here the ellipse
is obtained by stretching, rather than squashing, a circle.
In the simplest approach to the problem, one might assume that the ellipse's axes are parallel to
the sides of the square, so the constraint is simply that the major axis 2 r / cos  not exceed the
side 4r of the square. This gives cos   1 / 2 , or    / 3 .
However, a larger ellipse will fit in the square with the major axis along the diagonal. Rather
24 than inscribe a tilted ellipse inside the square, it may be easier to draw a tilted square around the
standard ellipse. The sides of the square, with slopes 1 , must then be tangent to the ellipse. To
achieve this we use implicit differentiation to locate the points where the ellipse has slope 1 ,
finding 2 x cos2   2 yy  0 with y   1 , so that y   x cos2  . (Of course, other methods of
finding the tangent lines, discussed above, lead to the same result.) Substituting back into the
equation of the ellipse gives the points of tangency as
r
( x, y ) 
( 1,  cos2  ),
2
cos  1  cos 
where all four sign combinations are possible. The x -intercept of the tangent line at the point
with positive coordinates can be computed as
r
1  cos2  .
cos 
The four tangent lines form a square whose diagonal is twice this value, so its side is this value
times 2 . The problem constraint is then
r 2
1  cos2   4r.
cos 
which can be solved to yield simply cos   1 / 7  0.37796. This translates into   67.79 o ,
versus 600 as found previously when the major axis was parallel to the sides.
Problem 15. A dog is tied up in a large field by a loop of rope running through its collar and
around two posts. If the posts are separated by a distance d , and the rope has length l , this
restricts the dog to an elliptical region in which the sum of focal distances is less than l  d . If a
straight river runs through the field along a given line, how close to this river can the dog get?
Consider a numerical example in which l  20 and d  6 . Let the ellipse be centered at
the origin with major axis along the x -axis, while the river has equation y  x  16 . Then
a  7, c  3, and the equation of the ellipse is ( x 2 / 49)  ( y 2 / 40)  1 . As in Problem ellipse9,
there are two tangent lines to the ellipse parallel to the river, namely y  x  89 . The minus
sign gives the tangent line nearer to the river, and the distance between this line and the river is
the closest the dog can get. The distance between parallel lines of slope m whose y -intercepts
differ by  b is b / m 2  1 , which in our case gives a distance of (16  89 ) / 2  4.6429 .
25 Unit 3: The Parabola
Focus WoT's and WoU's
 Understanding the way the equation of a locus changes depending on how it is placed
relative to the coordinate system, and that algebraic transformations induce a translation of
the locus.
 Structural way of thinking. Try to generalize established properties and explore a parameter
space (if an ellipse has 0  e  1 , what happens for e  1 ?).
 PGA way of thinking: Attending to interrelationships between physical/perceptual,
geometric, and algebraic realities.
 Algebraic invariance way of thinking. An equation can be rewritten in various forms which
make certain properties more noticeable. The link between the forms is provided by
something which remains invariant, for example the solution set.
 Usefulness of completing the square.
Classroom Problem 1
What locus satisfies the focus/directrix definition with 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, 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. Examples that were discussed in our class include:

Focus ( k ,0) , directrix y -axis: y 2  2kx  k 2 .


Focus ( k ,0) , directrix x   k : y 2  4kx .
Focus (0, k ) , directrix x -axis: x 2  2ky  k 2 , or y  ( x 2  k 2 ) / 2k .

Focus (k / 2,0) , directrix x   k / 2 : y 2  2kx .

Focus (0, 0) , directrix Ax  By  C  0 , 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 .
Students should see a need to reconcile these equations with their previous understanding of
a parabola, probably defined as the graph of a quadratic function y  ax 2  bx  c . Only one of
the above equations has this form, but students should recognize that reversing the roles of x
and y all but the last to be put into this standard form. The last example, with a general line as
directrix, creates Necessity for the formula for the distance from a point (focus) to a line (see line
unit).
One pair of students in our classroom interpreted this problem geometrically rather than
algebraically, and they presented a geometric construction of points on the locus. Given the focus
F and directrix l , choose a point Q on l , draw FQ and mark its midpoint M . Then the
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). The
required condition FP  QP follows from
FMP  QMP , by SAS. One can use this
construction to see that the locus is
symmetric about an axis through F and
normal to l . Another student translated this
construction directly into algebra. Taking F
to be (0, k ) and l to be the x -axis, let Q be
at ( x0 ,0) . One easily calculates that M is
( x0 / 2, k / 2) and P is at ( x, y ) where
x  x0 and y  ( x  k ) / 2k . Eliminating x0 yields y  ( x 2  k 2 ) / 2k , one of the previously
calculated equations for a parablola.
This connection between geometry and algebra stimulated a discussion of whether the
constructive approach has any pedagogical advantage compared to the direct derivation of the
algebraic equation from PF  Pl , which seems easier. A student suggests that it requires, and
shows, greater understanding of the construction and the definition of parabola. It promotes the
PGA WoT by showing that the connection between geometry and algebra can be made at a
variety of stages of a computation or levels of analysis. It also provides Necessity for confronting
the distinction between a Cartesian equation and a parametric representation of a locus. The
construction naturally gives a parametric representation, parametrized by x0 , the coordinate of
the arbitrary point Q . Coincidentally, this problem is so simple that the parametric
representation immediately reduces to the Cartesian version.
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. (They can
also interchange x and y to reflect about the
line y  x , but we postpone the use of
rotations of axes until later.) Students must
carefully reason through the process to avoid
subtle sign errors. First, it is necessary to
decide whether the translations will be active or passive: are we moving the locus on a fixed
coordinate system, or are we moving the coordinate axes while keeping the same locus? In our
classroom we chose the former. How do we modify the equation f ( x, y )  0 in order, say, to
move the curve a distance a in the x direction? One should avoid the notation x  x  a unless
its meaning has been thoroughly discussed and agreed upon. Does it mean that each point
P ( x, y ) on the curve is moved to the new location P( x  a , y ) , or does it mean that we modify
the equation by substituting x  a for each occurrence of x ? In fact, if we move each point as
stated, the new curve has equation f ( x  a, y )  0 . Repeated reasoning will be needed for
students to understand the source of the sign change. One possibility is to introduce
x  x  a, y   y for the coordinates of the moved point, so that f ( x, y )  f ( x  a , y ).
Classroom Problem 2: How does the
equation obtained for the parabola depend
on the placement of the focus and directrix
relative to the chosen coordinate system?
Which placement leads to the simplest
equation? Algebraically, how can the other
equations be transformed into this one?
27 Supplementary and Practice Problems:
Problem 1:
c. Let C be the set of points in the plane satisfying the equation f ( x, y )  0 . Let C  be the
set of points obtained by translating the points in C up 3 units and to the right 5 units.
What equation is satisfied by the points in C  ?
d. Let C be the set of points in the plane satisfying the equation f ( x, y )  0 . Let C  be the
set of points obtained by translating the points in C down 3 units and to the right 5
units. What equation is satisfied by the points in C  ?
e. Let C be the set of points in the plane satisfying the equation f ( x, y )  0 . Let C  be the
set of points obtained by translating the points in C up 3 units and to the left 5 units.
What equation is satisfied by the points in C  ?
f. Find an equation defining the parabola obtained by translating the parabola
y 2  2kx  0 down m units and to the left n units.
Problem 2: Find the focus and the directrix of the parabola y  2 x 2 .
Problem 3: Find the focus and the directrix of the parabola y  2 x 2  16 x  31 .
Problem 4: Give an example of an ellipse and a parabola with the same focus and directrix.
Problem 5: Find a tangent line to the curve 5 y 2  x  0 that is perpendicular to the line
20 x  y  1 .
Problem 6: Prove that an appropriate intersection of a plane and cone gives a parabola. How
does the Dandelin sphere construction for the parabola compare to that for the ellipse?
In addition to considering how to place the Dandelin sphere, students should also realize that the
double Dandelin sphere argument does not apply.
Problem 7: Show that the tangent line to a parabola at a point P makes equal angles with the
line joining P to the focus and the line through P perpendicular to the directrix.
Problem 8: (Putnam Exam) A dart is thrown at a square dartboard, with equal probability of
landing anywhere on it. What is the probability that it lands closer to the center than to any
edge?
Take the square to be bounded by the lines x  1 and y  1 , so that the center is at the origin
and the area is 4 . The locus of points equidistant from the center and the side y  1 is the
parabola having these as focus and directrix; its equation is y  (1  x 2 ) / 2 . Rotations by 90
degrees give the parabolas corresponding to the other sides. The dart lands closer to the center
than to any edge if it lands in the closed region bounded by these parabolas, roughly an octagon
with curved sides. Using the octagonal symmetry, the area is 8 times that of the segment in the
first quadrant bounded by the y -axis, the parabola y  (1  x 2 ) / 2 , and the line y  x . The
probability is obtained by dividing by 4 , the area of the square, and is

2 1
0
(1  2 x  x 2 )dx  (4 2  5) / 3  0.219.
Unit 4: The Hyperbola
Focus WoT's and WoU's
 Determining tangent lines and asymptotes.
 Referential symbolic reasoning: “simplifying” the equation of a locus by translation of axes.
 PGA WoT.
28 

Structural way of thinking. Try to generalize established properties and explore a parameter
space (if an ellipse has 0  e  1 , what happens for e  1 ?).
Usefulness of completing the square.
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, b, m, n so that the line y  mx  n is tangent
to the hyperbola ( x 2 / a 2 )  ( y 2 / b2 )  1 .
(b) Prove that the tangent line to a hyperbola bisects the angle between the focal radii.
(c) Find the angle between a hyperbola and an ellipse that share the same foci.
Pedagogical Considerations
By this point, students should pose this question
themselves. Based on their experience, our students
conjectured that the locus is a hyperbola, meaning a
curve with equation ( x 2 / a 2 )  ( y 2 / b2 )  1 . To verify
this, students suggested a coordinate system in which the focus F is at ( k ,0) and the directrix l
is along the y -axis. 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 .
We now wish to “simplify” this equation to a “symmetric” form like that of the ellipse, for
example making any symmetry of the locus obvious. The following computation is remarkable
in that no high school student would recognize it as simplification. The equation gets more
complicated at each step, and even the final form is about as long as the initial form. It counts as
simplification only because we possess the PGA Way of Thinking that translating the coordinate
axes and introducing new parameters for combinations of e and k will clarify the geometry of
the locus. Expanding, collecting terms, and completing the square in x leads to
k
k 2e 2
( e 2  1)( x  2 ) 2  y 2  2 .
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  e2k1 ) 2 y 2
 2  1.
a2
b
This tells us that there is a better choice of coordinates than the one we made: we should
transform x  x  e2k1 and y  y . The arrows mean that each point ( x, y ) on the curve should
be moved right to the location given by the shifted coordinates. The final form of the equation of
the hyperbola is indeed ( x 2 / a 2 )  ( y 2 / b2 )  1 . Its x -intercepts are (  a ,0) .
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  (b / a ) x 2  a 2 .
Perhaps it is monotonically increasing, like
x ? Can we prove this? What does it mean? A
29 student suggested that larger x 's give larger y 's. How can we tell algebra this meaning? One
possibility is that if x1  x2 ,then y ( x1 )  y ( x2 ) ; that is, (b / a ) x12  a 2  (b / a ) x22  a 2 . This is
true because the composition of the increasing functions x 2  a 2 and x is increasing. Using the
derivative to test for an increasing function would be overkill; calculus is not needed here.
We need to locate the focus and directrix of the hyperbola given by our final equation, which
have been lost in translation (bad pun). Since the shift right was by a / e , the directrix is now the
line x  a / e . Applying the same shift to the focus locates it at ( ae, 0) . Note that a / e  a  ae .
Because of the symmetry of the curve, there is a second focus at (  ae, 0) and a corresponding
directrix at x   a / e .
In our classroom we distributed a textbook derivation of the equation of the hyperbola to
our students for comparison with their solution to this problem. The textbook derivation employs
20/20 hindsight to situate the focus at ( ae, 0) and the directrix at x  a / e from the beginning,
miraculously resulting in the simple standard form of the equation. We contrasted this with our
implementation of the Necessity Principle and the PGA WoT. Students should see for
themselves the effect of various placements of the focus and directrix against the coordinate axes
and decide which they consider simplest.
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 line tangent to the
hyperbola at ( x1 , y1 ) has the equation
xx1 yy1
 2  1. (1.4)
a2
b
This can be rewritten
b2 x 
a2 
y  2 1  x  ,
a y1 
y1 
 a2 
showing the slope and x-intercept K  , 0  . The x x1 
intercept clearly approaches the origin as our point
Figure 9: Limiting behavior of tangent line to a moves to infinity; the form of the slope necessitates
x1
hyperbola examining the ratio
y1
. From the equation of the
hyperbola, one obtains
x1
a

1  (b / y1 ) 2 .
y1
b
Thus, as y1   , we must have x1   also, in such a way that x1 / y1 has the limiting
value  a / b . Therefore, the tangent line (1.4) has the limiting form y   (b / a ) x , and the
asymptotes are these lines through the origin (center). One can show also that the vertical (or
30 perpendicular) distance from the hyperbola to an asymptote goes to zero in the limit.
To see the three-dimensional origin of the asymptotes, take a plane cutting a cone in the
hyperbola, and a parallel plane through the vertex cutting it in a pair of lines. The hyperbola
meets every generator of the cone except for these two lines. They should be related to the
asymptotes, but unfortunately do not lie in the plane of the hyperbola. Projecting them into this
plane, and translating them to pass through the center, indeed gives the asymptotes.
A parallel treatment of the parabola shows that it has no asymptotes. (It also has no center.) The
tangent line to the parabola y 2  2kx at ( x1 , y1 ) has the equation yy1  k ( x  x1 ) , which can also
be written
k
y
y  x 1.
y1
2
In the limit y1   , the slope goes to zero and the intercept to infinity; the tangent line has no
limiting position.
Classroom Problem 3: (a) Find a condition on
a, b, m, n so that the line y  mx  n is tangent to
the hyperbola ( x 2 / a 2 )  ( y 2 / b2 )  1 .
(b) Prove that the tangent line to a hyperbola bisects
the angle between the focal radii.
(c) Find the angle between a hyperbola and an
ellipse that share the same foci.
8
6
4


2
P


K
-5
F1
F2
5
-2
This type of problem should be
straightforward by now. The tangency
condition in part (a) is a 2 m 2  n 2  b 2 .
The equal-angle property of the
hyperbola can be established in any of
the ways used for the ellipse. For part
(c), if an ellipse and a hyperbola with
the same foci intersect at P , then the
tangent line to the hyperbola at P
coincides with the normal to the
ellipse there, because both bisect the
same angle between the focal radii
(See Figure 10). Therefore the two
curves are orthogonal at P . Students
may not realize without reflection that
neither the ellipse nor the hyperbola is
completely determined by specifying
their foci; one more parameter such as
the eccentricity is required. Thus we
have a whole family of ellipses and
hyperbolas, and all are orthogonal
when they intersect!
-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 ( 5,0) and (5,0) and directrices the lines
31 x  16 / 5 .
Problem 2: Find the center, eccentricity, foci, directrices, and asymptotes of the hyperbola
x 2  2 x  4 y 2  3  0.
Problem 3: Let F1 and F2 be the foci of the hyperbola
x2
a2

y2
b2
 1 . Show that P is a point on
the hyperbola if and only if | PF1  PF2 | 2a.
Problem 4:
Prove that an appropriate intersection of a plane and cone gives a hyperbola. How does the
Dandelin sphere construction for the hyperbola compare to that for the ellipse?
Problem 5: Let the tangent line to a hyperbola at a point P meet the asymptotes at points A, B .
(a) Show that P is the midpoint of the segment AB .
(b) Show that the area of the triangle ABO , where O is the center of the hyperbola, is
independent of which point P was chosen.
Problem 6: The Greeks (Pappus) used conic sections to trisect angles. Reconstruct their
method, as follows.
(a) Given a line segment AB , show that the locus of all points P such that PBA  2PAB is
(a branch of) a hyperbola. Identify its focus, directrix, and eccentricity.
(b) An angle AOB can be trisected as follows. Assuming without loss of generality that
OA  OB , draw a circle with center O passing through A and B . Next draw the hyperbola
determined as above by segment AB . Let it cut the circle at a point P in the interior of the
angle. Show that 3POB  AOB .
32 Unit 5: The Line
Focus Ways of Thinking and Ways of Understanding
 PGA way of thinking
 Algebraic invariance way of thinking
The line is an object that is so familiar to students, both perceptually and mathematically, that
they are likely to take the algebraic characterization of a line for granted (Of course the equation
y  mx  b defines a line!). For this reason, we do not place a unit on the line at the beginning of
this module—students would not understand the questions being asked at that point. However,
the preceding processes of characterizing various conic sections should have prepared these
students to recognize that the following problems are indeed legitimate questions.
Classroom Problem 1: Find the equation of a line.
Classroom Problem 2: Show that if A and B are both non-zero, then the equation
Ax  By  C  0 defines a line.
Pedagogical Considerations
Here we list three approaches to the problem. The first
approach uses similar triangles. Consider a line l . In order to
tell algebra about the line l , we should choose a coordinate
system; we will choose this system in such a way that l is not
parallel to either of the coordinate axes. A line is determined by two points; call them P( x1 , y1 )
Classroom Problem 1: Find
the equation of a line.
and Q( x2 , y2 ) . Let R ( x, y ) be any other point on the line l . Without loss of generality, we may
assume that x1  x2  x . Let S and T be the points with coordinates ( x2 , y1 ) and ( x, y2 ) ,


respectively. Since RT is parallel to the y  axis and QT is parallel to the x  axis, QTR is a
right angle. Similarly, PSQ is a right angle. Moreover, QRT and PQS are corresponding


angles formed by the parallel lines RT and QS and the transversal l and are therefore
congruent. Thus QRT is similar to PQS by AA similarity. It follows that
QS
PS
RT
 QT
, that is:
y2  y1 y  y2

.
x2  x1 x  x2
Naming this common ratio m and solving for y , we find that y  mx  b , where b  y2  mx2 .
Students should be able to explain why generalizing this argument implies that any point ( x, y )
on the line satisfies the equation y  mx  b . They should also realize the reason for fixing two
points that define the line, which we denote by particular coordinates, and considering any other
point on the line, which we denote with general coordinates ( x, y ) .
A second approach is to suppose that the given line l is the perpendicular bisector of
some segment PQ . If the coordinates of P and Q are ( x1 , y1 ) and ( x2 , y2 ) , respectively, and
( x, y ) is a point on l , then since ( x, y ) is equidistant from P and Q , we have
( x  x1 ) 2  ( y  y1 ) 2  ( x  x2 ) 2  ( y  y2 ) 2 , or equivalently, Ax  By  C  0 , where
A  2 x2  2 x1 , B  2 y2  2 y1 , and C  x12  x22  y12  y22 .
A third approach (requiring some familiarity with the notion of betweenness; P~R~Q
33 means R is between P and Q) is to take the concept of distance as primary and define a line as

having distances sum. That is, given points P and Q, we define PQ as the union of 3 sets:
1. {R : R ~ Q ~ P & RQ  QP  RP}
2. {R : Q ~ R ~ P & RQ  RP  QP}
3. {R : Q ~ P ~ R & QP  PR  QR}
The necessity for all 3 of these sets can be established through overaching problem 8. If one
blindly applies the distance formula, the calculation is lengthy. However, we can choose (say)

the x-axis to be along the direction of PQ , in which case it is clear that R has the desired
properties iff it has y-coordinate of 0. From an expert perspective, this is really defining a line
through defining an axis. That is, we define a primary line as the set of all ordered pairs (x,0), or
y=0. From this, we can apply translation and rotation to obtain the general form of any line.
Classroom Problem 2: Show
that if A and B are both nonzero, then the equation
Ax  By  C  0 defines a line.
All of the approaches to the previous problem can be
reversed. For the similar triangles approach, suppose that
the points P( x1 , y1 ) , Q ( x2 , y2 ) , and R( x3 , y3 ) satisfy the
equation Ax  By  C  0 , where x1  x2  x3 . In order to

show that PR is a line, one must show that the measure of
PQR is  . As above, one may introduce points S ( x2 , y1 ) and T ( x3 , y2 ) , this time using SAS
similarity to conclude that QRT is similar to PQS , where the equality of the ratios
RT
QT
QS
PS
and
follows from the equations Axi  Byi  C  0 for i  1, 2,3 . The conclusion then follows
since PQS and RQT are complementary and SQT is a right angle by construction. 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.
For the second approach, consider the segment whose endpoints have coordinates (0, k )
2
2
and ( A, k  B ) where k  A B2 B 2 C . 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 , or equivalently, if and only if
Ax  By  C  0 .
The third approach relies on the converse of the Betweenness Theorem (namely, if P , Q
and R are distinct points such that PQ + QR = PR , then Q is between P and R ). If we
assume that ( xi , yi ) satisfy the equation Ax  By  C  0 , for i  1, 2,3 and that the points satisfy
the hypotheses of the theorem, then the result follows from a straightforward calculation.
Supplementary and Practice Problems:
Problem 1: In school you learned that a line in the plane has the equation y  mx  b . What
does this mean? Specifically, explain the meaning of the following two assertions:
a. A line in the x - y plane has the equation y  mx  b .
b. The equation y  mx  b represents a line.
Prove these two assertions.
34 Problem 2: Given two points in the plane, ( a , b ) and ( c, d ) , prove that the equation of the line
yb xa

.
that goes through these points is
d b ca
Problem 3: Show that the distance from the point ( x0 , y0 ) to the line Ax  By  C  0 is given
| Ax0  By0  C |
by
.
A2  B 2
35 Unit 6: Conic Sections and Plane Rotations
Focus WoT's and WoU's
 PGA WoT.
 Algebraic invariance WoT.
 Effect of placement of locus relative to coordinate system on equation of locus; algebraic
transformation inducing a rotation of the locus.
 Need for Structure: classifying the conic sections represented by general quadratic equations.
 Necessitating the concept of eigenvector by locating the vertices of a rotated ellipse.
An appropriate capstone to this module would be the use of rotations to classify the conics
represented by general quadratic equations in two variables. Ideally, we would like to connect
the discussion to quadratic forms and the eigenvalues and eigenvectors of linear transformations.
Because our students had limited linear algebra background, we were only able to take the first
step in this direction.
Classroom Problem 1: Recall that a parabola with focus at the origin and directrix the line
Ax  By  C  0, with A2  B 2  1 , has the equation
B 2 x 2  A2 y 2  2 ACx  2 BCy  2 ABxy  C 2 .
This should simplify to one of our standard forms by rotating the parabola so that its directrix is
vertical. How is such a rotation carried out algebraically?
Classroom Problem 2: Intrigued by the idea of obtaining an ellipse by squashing a circle, a
student investigates the effect of other simple (linear) transformations on the circle. Beginning
with the circle x 2  y 2  r 2 , she creates a new curve by plotting the new points
( xˆ , yˆ )  ( x  2 y , 2 y ) for each point ( x, y ) on the circle. 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. What sets of coefficients lead to the
same curve, up to a rigid motion? How do they determine its geometric parameters?
Pedagogical Considerations:
In general, the vector ( A, B ) is normal to the
Classroom Problem 1: Recall that a
directrix, 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, with
vertical. In fact, in our classroom, we limited
A2  B 2  1 , 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 .
the needed rotation angle is 45 . 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. A point ( x, y ) has polar
directrix is vertical. How is such a rotation
coordinates ( r ,  ) , which rotates to ( r,    ) .
carried out algebraically?
In Cartesian form this is
[ r cos(   ), r sin(   )] . Expressing this in
terms of the original Cartesian coordinates necessitates the angle sum/difference identities from
trigonometry (“When will I ever use this?”). The new coordinates are
r (cos  cos   sin  sin  , sin  cos   cos  sin  ),
or
36 ( x cos   y sin  , y cos   x sin  ).
Thus we need to re-express the equation of the parabola in terms of the new variables
xˆ  x cos   y sin  , yˆ  y cos   x sin  .
Referential symbolic reasoning can save a lot of work over the brute-force inversion of these
equations to give x, y in terms of xˆ , yˆ . One way is to observe that the inverse rotation must be
by angle  . Another is to note that with A2  B 2  1 and tan   B / A , we can take
A  cos  , 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 . Students who have carried out the brute-force
approach are quite likely to appreciate the benefit of referential symbolic reasoning.
The inverse of the coordinate transformation
is ( x, y )  ( xˆ  yˆ , yˆ / 2) , and this suffices to
give the equation of the new curve as
5
ˆˆ  yˆ 2  r 2 .
xˆ 2  2 xy
4
The question is whether this can be
recognized as a rotated ellipse, and what the
rotation angle is. One could rotate this
equation by an arbitrary angle  , and try to
determine  so that the equation simplifies.
Conversely, one could rotate the standard
ellipse equation and compare with the
example we have. Instead, we will determine the rotation angle in a more geometric way and
make the connection with eigenvalues.
Note incidentally that the coordinate transformation in matrix notation appears as X̂  TX ,
where
1 2
 1 1 
T 
, T 1  

.
0 2
0 1 / 2
The quadratic form appearing in the equation of the ellipse is Xˆ t AXˆ , where
1 
1
A  (T 1 ) t T 1  
.
 1 5 / 4 
A student who suspects that the curve is an ellipse should try to locate its vertices, 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). One plausible way to do this is by using Lagrange multipliers to
extremize xˆ 2  yˆ 2 subject to the equation as a constraint. This leads quickly to the equations
xˆ  yˆ


 xˆ 
  xˆ  (5 / 4) yˆ     yˆ  . (1.5)


 
which might be recognized as precisely the eigenvalue problem for the matrix A (not T ). The
eigenvalues are   (9  65) / 8 , approximately 2.133 and 0.117 . Taking the dot product of
the eigenvalue equations (1.5) with ( xˆ , 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,
a student investigates the effect of other
simple (linear) transformations on the circle.
Beginning with the circle x 2  y 2  r 2 , she
creates a new curve by plotting the new
points ( xˆ , yˆ )  ( x  2 y , 2 y ) for each point
( x, y ) on the circle. What curve does she
obtain? What equation could she have
graphed to obtain the new curve directly?
37 xˆ 2  yˆ 2 
r

,
so that the semiaxes should be r /  , namely 2.92r and 0.685r .
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. Using the gradient of the quadratic form to find the normal vector, one obtains 2 AXˆ , and
therefore the vertices can again be found from AXˆ   Xˆ .
Students will now conjecture that the curve is a rotated version of the ellipse
2
2
 x   y 
2

 
 r ,
 2.92   0.685 
with its axes along the eigenvectors of A . The major axis would be along the unit vector
(0.750, 0.662) , which makes an angle of about 0.723 radians with the x -axis. To verify their
conjecture, they could use the formulas for rotation of axes at this point. Alternatively, 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, namely the major axis. This would
produce the equation of the curve once again, up to roundoff errors.
We briefly discuss the simpler equation first. Once
again, students who are familar with diagonalizing
symmetric matrices and their associated quadratic
forms will have an easier time than those who are
not. Those without this background may again start
by locating the vertices of the (presumed) conic,
either as the points of maximum/minimum distance
from the center (the origin, by symmetry) or the
points where the position vector is normal to the
curve. The position vectors of these points are
eigenvectors of the symmetric matrix
B / 2
 A
M 
.
B /2 C 
The quadratic equation for the eigenvalues is
B 2  4 AC
2
  ( A  C ) 
 0.
4
From this we learn that the sum of the eigenvalues, 1  2 , is A  C (the trace of M ); 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. What sets of
coefficients lead to the same curve, up
to a rigid motion? How do they
determine its geometric parameters?
product of the eigenvalues, 12 , is AC  ( B / 2)2 (the determinant of M ); and the eigenvalues
themselves are


  A  C  ( A  C ) 2  B 2 / 2.
In particular, the eigenvalues are always real. The simpler equation 1 x 2  2 y 2  F  0
definitely represents a conic, and would lead to a matrix having the same trace and determinant.
However, at this point students really must use the formulas for rotation of axes and explicitly
transform the original equation into this simpler form. The type of conic depends on whether the
38 eigenvalues have the same or opposite sign; it is a hyperbola when B 2  4 AC  0 and an ellipse
when B 2  4 AC  0 . The semiaxes of the ellipse are | F / i | . In the remaining case
B 2  4 AC  0 , one eigenvalue vanishes and the corresponding semiaxis is undefined, so
something is wrong. In fact, the original quadratic form is then a perfect square, and the “conic”
is either empty or degenerate: a pair of lines. An equation of the given form cannot represent a
parabola, which is evident from its symmetry about the origin.
Students should attempt to reduce the more general equation to the less general one by
translation of coordinate axes to remove the linear terms. Replacing x by x  a , and y by
y  b , the linear terms cancel out if the linear equations
2 Aa  Bb  D,
Ba  2Cb  E
are satisfied. That is, the plan succeeds unless B 2  4 AC  0 . In that case the locus is not
symmetric about any origin. It is still (generally) possible to remove the xy term by rotating one
axis to lie along the eigenvector with the nonvanishing eigenvalue, and to remove the term linear
in either x or y by a translation. The resulting locus, if nondegenerate, is a parabola.
Supplementary and Practice Problems:
Problem 1. Identify the type of conic section defined by the given equation (none of the conics
are degenerate):
a. x 2  2 3 xy  3 y 2  2 3 x  2 y  0
b. x 2  4 xy  y 2  1
c. 73x 2  72 xy  52 y 2  30 x  40 y  75  0
d. Use rotation of axes to graph the conic sections in parts (a)-(c).
39 Overarching supplemental problems
Problem 1 Describe the possible intersections of a (double) cone with a plane. 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. The discussion can go in many
directions, 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, z  mx, m  0 , about the z -axis.
The various positions of the rotating line are called generators of the cone. The vertex of the cone
is then at the origin and its axis is along the z -axis. Its “elevation angle” is the inclination  of
the line, tan   m ; the vertex angle is   2  . A point of the line at height z sweeps out a
circle of radius | z / m | , which is therefore the cross-section of the cone by a horizontal plane at
this height. This definition of a cone is thus equivalent to the union of lines joining the vertex to
such a circle. Fixing z in the equation of the cone must give the equation of this circle, so the
cone's equation must be z 2  m 2 ( x 2  y 2 ) .
We have already seen that a horizontal plane cuts the cone in a circle (or a point if it passes
through the vertex). The xz plane meets it in the pair of lines z  mx making angle   2 
with each other, and any other plane containing the z -axis gives the same picture. 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  . For inclination angle    we get a single
generator of the cone. The case    may challenge students' visual intuition. A representative
example of such a plane is z  ax with a  m . Eliminating z between this and the equation of
the cone gives
a 2  m2
y  x
.
m2
Together with z  ax these are equations of two lines through the origin, two generators of the
cone. One possible geometric explanation is as follows. The plane will meet a circular crosssection of the cone in two points, and therefore it contains the two generators through these
points and no point of any other generator. Visually, as a rotating generator sweeps out the cone,
it passes from one side of the cutting plane to the other and back again. Therefore there are two
moments when it lies in the plane, by continuity.
Considering planes not containing the vertex, visual intuition may suggest that the intersections
are curves which are either closed and oval, or open and roughly parabolic. A direct algebraic
approach is challenging. For example, the plane z  ax  1 leads to
(m 2  a 2 ) x 2  m 2 y 2  2ax  1  0 . Students may observe that the relative size of m and a
determines the sign of the x 2 term, and that the case m  a does look parabolic. However, this is
the equation for the projection of the intersection curve into the xy plane, not its equation as a
plane curve in the cutting plane, so the interpretation is problematic. Eventually, students may
relate the coordinate x  in the cutting plane to the spatial Cartesian coordinate x by x  x  cos 
; for now they may continue intuitively. Imagine that an initially horizontal cutting plane slowly
rotates to a position with    . The circular intersection deforms into an oval shape and clearly
remains closed. In particular, the intersection curve contains a single point on each generator.
The deformation continues until    , when the cutting plane becomes parallel to a single
generator of the cone, which it no longer meets: that point of intersection moved off to infinity as
40 the oval stretched. The intersection curve is now open and unbounded. Continuing to    , 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. The three qualitatively
distinct intersection curves are called ellipses (from ellipsis, meaning deficiency or falling short),
parabolas (placing beside or comparable to), and hyperbolas (throwing beyond). (The names, due
to Appolonius, do not arise directly from the picture of slicing a cone.)
Problem 2 A sphere is illuminated by a point light source some distance away. Explain why its
shadow on a plane surface will be a conic section.
Draw the line through the light source and the center of the sphere. This will be the axis of a
cone. Rotate this line about the light source until it is tangent to the sphere. It is now a generator
of the cone. Rotating the generator about the axis produces the cone, which is tangent to the
sphere along a circle. The elevation angle is given by cos   R / D , where R is the radius of
the sphere and D is the distance from the light source to the center of the sphere. Now consider
the sphere's shadow on a plane placed behind it. Light rays from the source reach this plane if
they are external to the cone, and do not if internal. The boundary of the shadow is therefore the
intersection of the cone with the plane. Only the single nappe of the cone tangent to the sphere is
relevant; only one branch of a hyperbolic shadow appears. As mentioned, if the sphere is actually
in contact with (“resting on”) the plane, we call the point of contact a focus of the conic section.
Every (nondegenerate) conic section arises as the shadow of a sphere in contact with the cutting
plane, either by an intuitive argument in which one blows up a balloon enclosed between the
cone and plane until it is tangent to both, or by the fact, applied to a plane cross-section of the
diagram, that every triangle has an inscribed circle (applicable to the ellipse case) as well as three
excircles (in the other cases). 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. This is similar
in spirit to Problem circle1, whether a circle determines its center, and may necessitate a
characterization of a conic section as a locus in its plane.
Problem 3 We define the distance from a point to a line as the perpendicular distance. How
might we define the distance from a point to an ellipse (parabola, hyperbola)? Given a point not
on an ellipse (parabola, hyperbola), construct the point on the ellipse that realizes this distance.
Problem 4 Given two point A and B in the plane, 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, in any triangle, the following lines meet at a point.
a. The three medians (lines from a vertex to the midpoint of the opposite side)
b. The three altitudes (lines through a vertex perpendicular to the opposite side, extended
if necessary)
c. 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 .
a
b
41 Problem 7: Fix two points A and B in the plane. 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. If students are not careful with the distances, they may incorrectly answer that the
locus is a line, rather than merely the segment. This provides an opportunity to challenge their
intuitive assumption that an equation derived from a locus is equivalent to that locus.
Problem 8: A segment AB with endpoints on the sides of a right angle moves so that the
distance AB remains fixed. What is the locus of all the midpoints of AB ?
B
M
A
References
There are innumerable sources for information on conic sections. There follow a few which are
less well-known, or take an unusual approach.
W. H. Drew, A Geometrical Treatise on the Conic Sections, Macmillan 1875 .
J. K. Johnstone and C.-K. Shene, Computing the intersection of a plane and a natural quadric,
Computers and Graphics 16(2)(1992)179 -- 186 .
42 V.L. Hansen, Shadows of the Circle, World Scientific 1998 .
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