University of Waterloo
Faculty of Mathematics
Centre for Education in
Mathematics and Computing
Senior Math Circles
February 18, 2009
Conics III
Eccentricity of Conics
Fix a point F called the focus, a line d called the directrix, and a real number
e > 0 (the eccentricity).
Claim: The locus {P : kP F k = ekP dk}
for e = 1, and a hyperbola for e > 1.
is an ellipse for e < 1, parabola
Proof:
Use a similarity transformation to take d to the y-axis and F to (1, 0). Then
P = (x, y) is on the locus if and only if
p
(x − 1)2 + y 2 = e|x|
⇐⇒ (x − 1)2 + y 2 = e2 x2
2
, a parabola.
if e = 1 this gives −2x + 1 + y 2 = 0 or x = 1+y
2
Otherwise we can rewrite as
1
1
2
2
(1 − e2 )(x − 1−e
2 ) + y = 1−e2 − 1
which is an ellipse when 1 − e2 > 0, a hyperbola when 1 − e2 < 0. Reflective Properties
When light reflects off a curve, the angles of incidence and angles of reflection
are equal (with respect to the tangent):
This definition of reflection can be used to prove reflective properties of conics.
Proof of reflective property of ellipse:
(rays leaving one focus will pass through the other focus after reflecting):
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Let F1 , F2 be the foci and X any point on the ellipse.
If F1 → X → F2 is indeed the light’s path, then
the tangent must be the extended angle bisector l of
∠F1 XF2 and vice-versa.
F1
X
l
F2
Suppose, for the sake of contradiction, that l is not tangent to the ellipse. Then
it hits the ellipse twice. Let X 0 be a point on l but strictly inside the ellipse; then
kF1 X 0 k + kF2 X 0 k < 2a where a is the focal radius.
On the other hand, construct the image of F2 after reflecting through l; call
is F20 . Since F1 XF20 is a straight line and F1 X 0 F20 is not, the triangle inequality
gives:
kF1 X 0 k+kF2 X 0 k = kF1 X 0 k+kF20 X 0 k > kF1 Xk+kF20 Xk = kF1 Xk+kF2 Xk = 2a.
Since kF1 X 0 k + kF2 X 0 k cannot be both larger than and smaller than 2a, we obtain the desired contradiction. More generally, for conics, A → B → C, with B on the conic, is a valid reflection if B is a local minimizer or maximizer of kABk + kBCk. This approach
can be used to prove the reflective property of ellipses and hyperbolas.
Envelope
Envelope problems are similar to locus problems, except we are given a moving
line instead of a moving point . . .
A family of lines is said to envelop a curve if each line in the family is tangent to the curves;
the curve is called the envelope of the family.
Duality between ellipse and its set of tangents:
set of tangents
envelope
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Despite the apparent grossness, envelope problems are often solvable by simple algebraic methods.
Key observation: any point on the envelope lies in exactly one member of the
family of lines.
Sample Problem: Consider the family of lines whose x- and y-intercept add
up to 1. Find the envelope of this family.
Solution:
Let t denote a parameter which we will use to describe the family of lines; specifically, let any line in this family have x-intercept (t, 0) and y-intercept (0, 1 − t).
This line has equation
y
x
+ 1−t
= 1. (*)
t
Now we use the “key observation”: for any point x, y on the curve, there is only
one t for which (*) holds.
Solve (*) for t:
(1 − t)x + ty = t(1 − t)
⇐⇒ t2 + t(y − x − 1) + x = 0,
a quadratic in t.
This has one root if and only if the discriminant “B 2 − 4AC” vanishes, i.e.
(y − x − 1)2 − 4x = 0. Rule. Let αx2 + βxy + δx + y + η = 0 be a conic. It is a

2

β − 4αδ < 0
ellipse
β 2 − 4αδ = 0
parabola
if

 2
β − 4αδ > 0
hyperbola
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Ellipse
Parabola
F1
b
b
Picture
b
Typical
Equations
Typical
Definition
F1
F
d
F2
Hyperbola
x2 + y 2 = 1,
y − x2 = 0,
x2 /a + y 2 /b = c,
y = ax2 + bx + c,
(x−x0 )2 /a+(y−y0)2 /b = c
x = ay 2 + by + c
{P : ||P F1 || + ||P F2 || = 2a}
{P : ||P F || = ||P d||}
b
b
F2
x2 − y 2 = 1,
x2 /a − y 2 /b = c,
(x−x0 )2 /a−(y−y0)2 /b = c,
xy = c, (x−x0 )(y−y0 ) = c
{P : ||P F1 || − ||P F2 || = ±2a}
Rays leaving F1 , after reRays leaving F , after reRays leaving F1 , after reReflective
flecting off the hyperbola,
flecting off the parabola,
flecting off the ellipse, hit
Property
travel directly away from
are perpendicular to d.
F2 .
F2 .
0<e<1
e=1
e>1
Eccentricity*
*: This gives an alternate way of defining ellipses and hyperbolas. Namely, for a line d and point F ,
these curves can be defined to be the locus {P : ||P F || = e||P d||}.
Obtaining as a Conic Section
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Exercises
1. Find an equation of the envelope of the family of lines whose x- and y-intercepts have
a constant product k.
2. Find an equation of the envelope of the family of circles whose centers lie on the
parabola {y = x2 } and who are tangent to the x-axis.
3. The midpoint of a chord of the circle x2 + y 2 = r2 is on a fixed straight line x = a,
0 < a < r. Find an equation of the envelope of the family of such chords. (Hint: let
the midpoint be (b, a) & consider the chord’s slope).
4. Let P be any fixed point inside a circle C
with centre O and radius r. Let A be a
variable point on the circle. Let l be the
perpendicular to AP through A. In this
problem we find the envelope of l as A varies.
(note: you can use a “set square” or any
rectangular object to sketch the envelope.)
l
A
P
O
(a) Let P 0 be the image of P after reflecting P through O, and let Pb be the image of
P 0 after reflecting P 0 through l. Show kP Pbk = 2r (Hint: inscribe a rectangle in
C such that one side is l).
(b) Show that exactly one point X on l satisfies kP Xk + kP 0 Xk = 2r, and that the
rest satisfy kP Xk + kP 0 Xk > 2r.
(c) Determine, with proof, the envelope of l as A varies.
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