55. The Curvature of Conic Sections
Determine the curvature of a conic section.
The answer is remarkably simple: let P be a point on the conic, n the length of the
normal from P to the axis and p half the length of the latus rectum. (The LR is the line
segment through a focus perpendicular to the axis.) Then the radius of curvature at P is
and the curvature at P is κ =
p2
n3
n3
p2
.
directrix
tangent
P
n
axis
F
tangent
P
n
F2
F1
P
F1
n
F2
tangent
Let the conic section by c, its latus rectum have length 2p (some books say the
parameter is 2p), e its eccentricity. An equation for c is
qx 2 + y 2 − 2px = 0 with q = 1 − e 2 .
(q = 0 ⇒ y 2 = 2px and this is the equation of a parabola with LR of length 2p. If q ≠ 0,
complete the square in the x terms to get the equation of an ellipse or hyperbola with
eccentricity e and LR of length 2p. )
1
p
2p = 1.99
p = 1.00 cm
L R = 1.99 cm
8
e
e = 0.90
6
q = 0.20
4
2
-5
F
5
10
15
F
5
10
15
F
5
10
15
-2
-4
-6
p
2p = 1.99
p = 1.00 cm
L R = 1.99 cm
8
e
e = 1.00
6
q = 0.00
4
2
-5
-2
-4
-6
p
2p = 1.99
p = 1.00 cm
L R = 1.99 cm
8
e
e = 1.23
6
q = -0.51
4
2
-5
-2
-4
-6
Definition Let Px, y be any point in the plane on or outside c, i.e., for which
qx 2 + y 2 − 2px ≥ 0, and line AB through P be a secant line of c. Let θ be the directed
2
or signed angle from vector ⟨1, 0⟩, i.e., the vector from 0, 0 to 1, 0 to vector PA.
p
p = 1.01 cm
2p = 2.01 Latu s rectum = 2.01 cm
e
8
e = 0.90
q = 0.18
6
4
P(x,y)
A
2
θ
F
-5
5
10
15
-2
B
-4
-6
We define
1.
2.
3.
D = D θ = sin 2 θ + q cos 2 θ; D is the direction number for θ;
E = E P,θ = 2u cos θ − y sin θ where u = p − qx; E is the emanant at P in
direction θ;
F = F P = qx 2 + y 2 − 2px; F is the power of P with respect to c.
Then with the notation above,
1.
2.
E = D ⋅ PA + PB and
F = D ⋅ PA ⋅ PB.
Proof. A = x + PA cos θ, y + PA sin θ, and B = x + PB cos θ, y + PB sin θ. Since A and B
are on c, their coordinates satisfy the equation qx 2 + y 2 − 2px = 0, and we get (after
substitution and simplification) D ⋅ PA 2 − E ⋅ PA + F = 0 and D ⋅ PB 2 − E ⋅ PB + F = 0.
PA and PB are thus roots or the quadratic equation Du 2 − Eu + F = 0, and the
conclusion follows.
□
Now to find the curvature at a point Px, y on c. Let T be the tangent line at P, making
an angle ϕ with the x-axis. Then
D ϕ = sin 2 ϕ + q cos 2 ϕ
=
u2
n2
=
p2
n2
y2
+ q n2
since u = p − qx is the subnormal:
3
8
6
T
Q
4
t
A
2
P(x,y)
n
ϕ
θ
subnormal
p-qx
-5
-2
5
10
15
B
-4
-6
and u 2 + qy 2 = p − qx 2 + qy 2 = qqx 2 + y 2 − 2px + p 2 = p 2 . The inward pointing normal at P
makes a directed angle of ϕ − 90 ∘ = θ with the positive x-axis, so
D θ = sin 2 θ + q cos 2 θ = cos 2 ϕ + q sin 2 ϕ
and
E θ = 2u cos θ − y sin θ = 2y cos ϕ + u sin ϕ = 2n.
The normal line at P intersects c at another point, at a distance, say from P.
8
T
6
Q
4
t
2
P(x,y)
ϕ
-5
N
5
10
15
-2
-4
-6
It follows that E θ = D θ ⋅ 0 +  or
(1)
2n = D θ ⋅ .
Let Q be on tangent line T at a distance of t from P. Let line QA be a secant to c,
perpendicular to T, intersecting c again at B with S = QB > QA = s :
4
8
6
T
4
Q
t
P(x,y)
2
A
ϕ
-5
5
N
10
15
-2
B
QA=s, QB=S > s.
-4
-6
Since D ϕ−180 ∘ = D ϕ , each of the following represents the power of Q with respect to c :
D ϕ ⋅ t 2 = D θ ⋅ s ⋅ S.
(2)
Now let k be a circle with center on the internal normal, radius ρ, tangent to c at P, and
intersecting line QA at C and D with s 0 = QC < QD = S 0 :
8
6
T
4
P(x,y)
2
t
Q
A
C
ϕ
-5
5
10
15
D
-2
k
B
-4
QC = s0, QD = S0, S0 > s0
-6
Considering only the circle k, tangent PQ and secant QC, we have
(3)
t2 = S0s0.
Divide (2) by (3) to get D ϕ S 0 s 0 = D θ Ss; multiply by and use (1) to get
D ϕ S 0 s 0 = D θ Ss = 2nSs. This last equation can be written as
5
s
S0 Dϕρ
s 0 = S ⋅ 2ρ ⋅ n .
As Q → P along the tangent line T, S → and S 0 → 2ρ. If k is the circle of curvature, then
s 0 → s as Q → P, i.e., k most nearly approximates c in the vicinity of P. Then D ϕ ρ → n, and
3
ρ = Dnϕ = p 2n/n 2 = np 2 is the radius of curvature.
Note 1. Let γ be the undirected angle between the normal at P and the focal radius FP.
p
p = 1.00 cm
2p = 1.99
L R = 1.99 cm
e = 0.90
e
8
6
q = 0.19
4
P(x,y)
2
γ
H
F
-5
5
K
10
15
Z
-2
-4
-6
We will momentarily show that cos γ =
p
n
. Then the radius of curvature is ρ =
n
cos 2 γ
,
and this is PZ in the figure above, since
PZ
PK
=
PK
PH
PZ
=
PK 2
PH
=
n 2 sec 2 γ
n
.
The center of curvature Z is thus easy to construct.
Note 2. By the law of cosines with △FHP and FP = r and FH = w, we have
2 2
2
w 2 = n 2 + r 2 − 2nr cos γ and cos γ = n +r2nr−w . Let the focus F have coordinates
p
a, 0 = 1+e , 0 .
6
p
p = 1.01 cm
2p = 2.01
8
Latu s rectum = 2.0 1 cm
e
6
e = 0.90
qx2 + y 2 - 2px = 0
q = 0.19
4
P(x,y)
2
n
r
F
-5
H
5
10
15
-2
F = (a,0),
u = p-qx is the subnormal,
w = FH = x - a + u.
-4
-6
First we find r :
r 2 = x − a 2 + y 2
= x 2 − 2ax + a 2 + 2px − qx 2
= e 2 x 2 + 2xp − a + a 2
= e 2 x 2 + 2xea + a 2
= ex + a 2
and r = ex + a. Then
n 2 + r 2 − w 2 = y 2 + u 2  + y 2 + x − a 2 − x − a + u 2
= 2y 2 − ux − a
= 22px − qx 2 − p − qxx − a
= 2px + pa − aqx
= 2 px + pa −
p
1+e
1 − e 2 x
= 2px + pa − p1 − ex
= 2pex + a
= 2pr,
and it follows that cos γ =
2pr
2nr
=
p
n
.
Note 3. At P = 0, 0 (or another vertex if c is an ellipse or hyperbola), the normal is the
line y = 0, so n is undefined, however, lim n = p, and the radius of curvature there
P→0,0
is p.
7
Note 4. It’s a nice exercise to use a geometry software program to construct the circle of
curvature to a conic.
y ′′
Note 5. A twice-differentiable function y of x has curvature κ =
Calculus). With y 2 = 2px − qx 2 , 1 + y ′  2 =
after some algebra, and we get κ =
p2
n3
n2
y2
1+y ′ 
2
3
2
at x, y (by
p2
and y ′′ = ± y 3 (− if y > 0 and + if y < 0
.
8