Under which conditions do ellipsoids have a focal property?

Question

Given an ellipsoid $E$, we consider the trajectories of light inside $E$ assuming that $\partial E$ would be a mirror. In other words, let a light trajectory be piecewise linear path $\gamma:[0,\infty)\rightarrow E$ such that that at each $t$ such that $\gamma(t)\in\partial E$, the tangent vector $v=\gamma'(t)$ changes to its symmetric reflection with respect the hyperplane $T_{\gamma(t)}\partial E$.

Making the above concrete, assume that, after a proper orthogonal change of variables, our ellipsoid is of the form $$ E:=\{x\in\mathbb{R}^n\mid \sum_{i=1}^n\lambda_ix_i^2\leq 1\} $$ with $0<\lambda_1\leq\lambda_2\leq\cdots\leq\lambda_n$. Then in a light trajectory, the tangent vector $v$ at $x\in\partial E$ would change from $v$ to $$ \left(\mathbb{I}-2\frac{\Lambda xx^T\Lambda}{x^T\Lambda^2x}\right)v. $$ where $\Lambda$ is the diagonal matrix whose diagonal is $\lambda:=(\lambda_1,\ldots,\lambda_n)$.

In the particular case of the ellipse, i.e., a 2-ellipsoid, it is well-known the focal property by which any light trajectory that starts at one of the foci returns to one of the foci exactly after one reflection.

An easy generalization to three dimensions is to consider the revolution ellipsoid obtain by rotating the ellipse around the axis containing the foci, which correspond to the case $n=3$, $\lambda_1<\lambda_2=\lambda_3$. In this case, the symmetry allows one to see that this 3-ellipsoid has a focal property.

However, if instead of rotating with respect the axis containing the foci, we rotate with respect the other foci, things get a priori more complicated. This is the case $n=3$, $\lambda_1=\lambda_2< \lambda_3$. In this case, the obtained 3-ellipsoid does not have a finite number of 'foci', but we get a circle $C$ of them.

The main question is: Does the above ellipsoid have a focal property with respect $C$? I.e., does every light trajectory starting in $C$ return to $C$ after a reflection (or maybe some fixed number of reflections)? If not, is it the statement true if we substitute $C$ by its convex hull?

A more general question is: does the focal property of the ellipse generalise (under maybe some hypothesis) to higher dimensional ellipsoids?

[This was a question made to me by an experimental physicist, but after thinking about it I have been unable to find any reference about it despite its elementary looking form].

EDIT: Following the comment of @Fly by Night, I made the question more precise.

Answer 1

A billiard trajectory inside an ellipsoid in $n$-dimensional space is tangent to $n-1$ quadrics confocal with this ellipsoid.

In your case, the disk bounded by $C$ is a limit of ellipsoids confocal to $E$, lines intersecting $C$ play the role of tangents, so a line starting at $C$ returns to $C$ after one reflection.

The proof of the general result above can be found in

Tabachnikov, Serge, Geometry and billiards, Student Mathematical Library 30. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3919-5/pbk). xi, 176 p. (2005). ZBL1119.37001.

The idea is to use ellipsoidal coordinates as Jacobi did when studying geodesics on ellipsoids. Billiard trajectory inside an ellipsoid is the limit case of a geodesic on an ellipsoid one dimension higher (as the higher-dimensional ellipsoid flattens, a geodesic going over the "edge" becomes billiard trajectory).

The argument in the book cited is for ellipsoids with different half-axes. The general case is proved by going to the limit. As two of $\lambda_i$ approach, some of the quadrics from confocal family degenerate (to double planes, I guess, so that a trajectory whose first segment lies in such a plane, always remains in the plane). But here we are interested only in those quadrics from the family which are ellipsoids. Let $\lambda_i(t)$ be all distinct for $t \ne 0$ and some of them coincide for $t=0$. For every $t$ consider the corresponding ellipsoid $E(t)$ and its confocal family. The billiard trajectories inside $E(t)$ depend continuously on $t$, and so do the ellipsoids confocal to $E(t)$. Thus one can take limit as $t \to 0$.