QUESTION: Let $E$ be an ellipse with centre at origin $O$ and the major and minor axis to be $2a$ and $2b$ respectively. Let $\theta$ be the acute angle at which $E$ is cut by a circle with centre at the origin. Prove that the maximum possible value of $\theta$ is $tan^{-1}(\frac{a^2-b^2}{2ab})$
MY ANSWER: I tried to use parametric coordinates for this one, but I couldn't end up with the proof. I wrote an expression of the acute angle $(tan\theta=\frac{m_1-m_2}{1+m_1m_2})$ (where $m_1$ and $m_2$ are the slopes of the circle and the ellipse respectively at the point of consideration). Then I tried to differentiate the expression, and set it to zero. But there are too many variables popping in, and this process does not seem to be of any help. Is there a smarter way to think about it?
Thank you.
Use the standard parameterization $(a\cos t,b\sin t)$. The normal to the ellipse at this point has direction vector $\left(b\cos t,a\sin t\right)$, while a normal to the circle is simply the position vector of the point itself. We then have $$m_1 = {a\sin t\over b\cos t}, m_2 = {b\sin t\over a\cos t}$$ and upon plugging this into the formula for the tangent of the difference of two angles that you have in your question and simplifying, you’ll end up with a positive constant times $\sin{2t}$. The maximum value of that expression should be obvious. You can use tangent vectors instead of normals if you like—the ensuing calculation is similar.