How do I deal with reflections inside an ellipse?

Question

Suppose I have an ellipse with foci $F_1$ and $F_2$. How do I show that any ray of light which intersects the segment connecting the foci will have subsequent reflections that always are tangent to the/a hyperbola formed with the same foci?

Answer 1

Elliptical billards are a classical topic. For a quick introduction see:

Billiards, by Serge Tabachnikov; Panoramas et Synthèses 1, SMF, Paris 1995 (in English),

p. 28ff., in particular Corollary 2.3.4 on p. 32.

Answer 2

Here's an interesting observation that fell out of my calculations: a "construction" of the tangent hyperbola (or ellipse) for a given chord.

Let $\overleftrightarrow{AB}$ be an extended chord of ellipse $ABCD$ passing between the foci, and let $P$ be the pole of $\overleftrightarrow{AB}$ (that is, the intersection of the tangents at $A$ and $B$). Drop a perpendicular from $P$ onto $\overleftrightarrow{AB}$ at $F$. Then the confocal hyperbola through $F$ is tangent to $\overleftrightarrow{AB}$ there.

(When the chord (see $\overleftrightarrow{CD}$) doesn't pass through the foci, dropping a perpendicular from the pole ($Q$) gives the point ($G$) of tangency on the confocal ellipse tangent to the chord.)

For proof, one can do what I did: slog through trigonometric simplifications based on the standard parameterization of the ellipse, using the fact that the tangent vector to point $(a\cos\theta, b\sin\theta)$ is $(-a\sin\theta,b\cos\theta)$. (A computer algebra system like Mathematica really helps here!) One could also seek a more-satisfying geometric argument.

This observation reduces the original problem to showing that consecutive segments in a reflection path (say, $BAB^\prime$) give rise to points ($F$ and $F^\prime$) on the same hyperbola.

Again, an arduous trigonometric-coordinate argument will do the trick. Tantalizingly, the poles ($P$ and $P^\prime$) in this scenario both lie on the tangent at $A$, and by the reflection property, we have $\triangle APF\sim\triangle AP^\prime F^\prime$. As mentioned in my comment to @ChristianBlatter's answer, Serge Tabachnikov's "Geometry and Billiards" has a concise geometric proof; even so, the above may lead to a not-unreasonable alternative.

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Edit. I've been playing with this a bit more, and found the following:

Let the ellipse have focal radius $c$ and major radius $a$ (and minor radius $b := \sqrt{a^2-c^2}$). Using the standard parameterization $E(\theta) = (a\cos\theta, b\sin\theta)$, take chord $\overline{AB}$ to have endpoints $A=E(2\alpha)$ and $B=E(2\beta)$. Let $d$ be the transverse/major radius of the confocal conic tangent to $\overline{AB}$ (according as the conic is a hyperbola/ellipse). Then,
$$\frac{d^2}{a^2} = \frac{ a^2 - c^2 \cos^2(\alpha+\beta) }{ a^2 \cos^2(\alpha-\beta) - c^2 \cos 2\alpha \cos 2\beta } \qquad (\star)$$

That all chords of a light path are tangent to a particular confocal conic implies that the left-hand side of $(\star)$ ---and therefore also the right-hand side--- is constant over all chords of the light path.

We can improve $(\star)$ a bit by re-parameterizing the ellipse as a polar graph with a focus at the origin: $$r(\phi) = \frac{a^2-c^2}{a-c\cos\phi}$$ Here, we write $A = r(2\alpha^\star)$ and $B = r(2\beta^\star)$, and get

$$\frac{d^2}{a^2} = \frac{ c^2 - 2 a c \cos(\alpha^\star - \beta^\star) \cos(\alpha^\star + \beta^\star) + a^2 \cos^2(\alpha^\star - \beta^\star) }{ a^2 - 2 a c \cos(\alpha^\star - \beta^\star) \cos(\alpha^\star + \beta^\star) + c^2 \cos^2(\alpha^\star - \beta^\star) } \qquad(\star\star)$$

While $(\star\star)$ is longer than $(\star)$, it has more symmetry. Moreover, its geometric meaning is more clear:

• The numerator is the square of the length of the side of a triangle whose other sides, of length $c$ and $a\cos(\alpha^\star - \beta^\star)$, enclose an angle of measure $\alpha^\star + \beta^\star$.
• The denominator is the square of the length of the side of a triangle whose other sides, of length $a$ and $c\cos(\alpha^\star - \beta^\star)$, enclose an angle of measure $\alpha^\star + \beta^\star$.

One can readily construct representative segments based on the bisector of $\angle AOB$, although I personally can't "see" (yet?) in the construction why the ratio of their lengths should be a light-path constant.

Anyway, this result is probably well-known in the vast elliptical-billiards literature, but I didn't happen to come across it with a cursory scan of Tabachnikov's article, so I thought I'd share.