# Ellipse curvature

Let $$a,b>0$$ and let $$\alpha$$ be the ellipse $$\alpha(t)=\big(x(t),y(t)\big)=\big( a\cos t, b\sin t\big)$$.

Let $$(u(t),0)$$ be the point where the line through $$\alpha(t)$$ perpendicular to $$\alpha'(t)$$ meets the $$x$$-axis, and let $$d(t)$$ be the distance between $$\alpha(t)$$ and $$(u(t),0)$$. Find a formula for $$\kappa(t)$$ in terms of $$a$$, $$b$$ and $$d(t)$$.

The curvature of the ellipse has been discussed thoroughly in this question: how to calculate the curvature of an ellipse

However, I can't find a way to use $$d(t)$$ in expressing curvature.

I tried to sketch the graph and realized that I can inscribe a circle inside the ellipse. But that is as far as I can go.

I also tried to explicitly write out $$d(t)$$, but it gives me an ugly expression.

• What did you get for $d(t)$? You shold recognize some similar part as in $\kappa(t)$.
– WimC
Commented Jan 26, 2021 at 6:08
• I found that I made a calculation error. That is the reason that I am getting weird values and unable to recognize the pattern. Commented Jan 26, 2021 at 7:35

Equation of tangent

$$0=T(x,y)\equiv \frac{x\cos t}{a}+\frac{y\sin t}{b}-1$$

Equation of normal

$$0=N(x,y) \equiv \frac{ax}{\cos t}-\frac{by}{\sin t}-(a^2-b^2)$$

Note that $$N(u,0)=0$$ and

\begin{align} u(t) &= \frac{(a^2-b^2)\cos t}{a} \\ d(t) &= \frac{|T(u,0)|}{\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\ &= \frac{\left| -\sin^2 t-\dfrac{b^2\cos^2 t}{a^2} \right|} {\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}} \\ &= b^2\sqrt{\frac{\cos^2 t}{a^2}+\frac{\sin^2 t}{b^2}} \\ \kappa &= \frac{1}{a^2 b^2 \left( \dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2} \right)^{3/2}} \\ &= \frac{b^4}{a^2 d^3} \end{align}

• Could you explain how you got the formula for the distance?
– John
Commented Feb 1, 2021 at 5:38
• The distance of a point $(x',y')$ from the line $Ax+By+C=0$ is given by $$\left| \frac{Ax'+By'+C}{\sqrt{A^2+B^2}} \right|$$ and see the $3$D analogy in another post here. Commented Feb 1, 2021 at 5:49
• I guess what I meant is why is the distance between $\alpha(t)$ and $(u(t), 0)$ the same thing as the distance between the tangent and $(u(t), 0)$.
– John
Commented Feb 1, 2021 at 5:53
• Back to your post, $\alpha$ is the curve, $\alpha'$ is the tangent line, i.e. $T=0$. A line through $\alpha$ and perpendicular to $\alpha'$ is the normal through $\vec{r}=\alpha(t)$, i.e. $N=0$. The line cut the $x$-axis at $(u,0) \implies N(u,0)=0$. Commented Feb 1, 2021 at 6:01
• Strictly speaking, $\alpha'(t)$ is a tangent vector. In the language of vectors, tangent line should be $\vec{r}(\lambda;t)=\alpha(t)+\lambda \alpha'(t)$, whereas in coordinate geometry it reads $T(x,y)=0$. Commented Feb 1, 2021 at 6:08

It is well known that (see figure below, $$F$$ and $$G$$ are the foci):

$$\kappa={\cos\alpha\over2}\left({1\over p}+{1\over q}\right) ={a\cos\alpha\over pq}.$$ On the other hand, from the formula for the length of a bisector we get: $$d^2={b^2\over a^2}pq$$ while from the cosine rule applied to triangle $$FPG$$ we obtain $$\cos^2\alpha={b^2\over pq}.$$ Inserting these results into the formula for $$\kappa$$ we finally get: $$\kappa={b^4\over a^2d^3}.$$