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

