Ellipse curvature

Question

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.

Answer 1

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}

Answer 2

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: $$ \kappa={b^4\over a^2d^3}. $$