Maximize the distance between a line normal to an ellipse and its center My friend sent me this problem, which (upon Googling) seems to be from a Cornell class (1220?). Anywho. 

My advice to him was to parametrize the ellipse (say, in the first quadrant) with
$x = a \cos(t); y = b \sin(t)$, find normal lines, then use the formula for the distance between a point and a line. But then I started wondering: 

Is there a better way? (Also, will this method even lead to a solution?!?)


The same maximal distance occurs in each quadrant, so we can restrict attention to $t\in[0,\pi/2]$. The tangent vector at $t$ is $(-a\sin t,b\cos t)$. This vector is normal to the line, so we just have to take the scalar product of a unit vector in this direction with the position vector in order to find the distance of the origin from the line:
$$
\begin{eqnarray}
D
&=&
\left\lvert\frac{(-a\sin t,b\cos t)}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\cdot(a\cos t,b\sin t)\right\rvert
\\
&=&
\frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\;.
\end{eqnarray}$$
Differentiating with respect to $t$ yields
$$\frac{a^2\sin^4 t-b^2\cos^4t}{\left(a^2\sin^2t+b^2\cos^2t\right)^{3/2}}\;,$$
and setting this to zero yields
$$a^2\sin^4t=b^2\cos^4t\;,$$
$$t=\arctan\sqrt{\frac ba}\;.$$
Using $\cos t=1/\sqrt{1+\tan^2 t}$, we can evaluate $D$ at this parameter:
$$
\begin{eqnarray}
D
&=&
\frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}
\\
&=&
\frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\cos t
\\
&=&
\frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\frac1{\sqrt{1+\tan^2 t}}
\\
&=&
\frac{(a^2-b^2)\sqrt{b/a}}{\sqrt{a^2(b/a)+b^2}}\frac1{\sqrt{1+b/a}}
\\
&=&
\frac{a^2-b^2}{a+b}\;.
\\
&=&
a-b\;.
\end{eqnarray}$$
The result obviously supports your idea that there might be a simpler way to do this.
