# Will a ball fired through a focus of an ellipse eventually tend to a horizontal line?

A couple of years ago I came across this phenomenon which appears to be true although I am having difficulty proving it.

F and F' are foci of a billiard table in the shape of an ellipse. A ball is fired through one of the foci, I can prove that it will subsequently pass through the other focus (this is in fact the question that sparked off this idea). Now if we continue to follow the path of the ball then it seeems as though it will eventually tend to travel in a horizontal line. Some ideas I had were to create triangles and develop a recursive sequence of an angle and see if this tended to 0 or do a similar thing with the gradient of the path of the ball and see if this tended to 0 but couldn't quite get a nice formula either way.

This question is actually Exercise 4.3 in the following notes http://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf

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I took the liberty of adding an image (from math.cornell.edu/~mec/Summer2009/Remus/lesson2.html). –  Joseph O'Rourke Mar 21 '11 at 15:58
(Re-added the image, as the link was broken.) –  Joseph O'Rourke Jul 21 '13 at 19:54

Just a couple of hints. Let's say our ellipse has foci $F_1=(-a,0)$ and $F_2=(a,0)$. Your idea of looking at the sequence of angles $\theta_n$ that the ray makes with the x-axis after it passes from $F_1$ for the $n$th time, can be made to work!

At every stage consider $F_1\to X\to F_2\to Y\to F_1\to X'$, where $X,Y,X'$ are on the boundary. If $F_1X$ makes an angle $\theta_n$ with the x-axis then, $F_1X'$ makes an angle of $\theta_{n+1}$, however simple geometry shows $$\theta_{n+1}=\angle X'F_1F_2=\pi-\angle YF_1F_2=\angle F_2F_1X+(\angle F_1XF_2+\angle F_2YF_1)$$ $$\geq \angle F_2F_1X=\theta_n$$

So $\{\theta_n\}$ is a monotonic increasing sequence on $(0,\pi)$, and therefore it must have a limit. It is easy to show that this limit has to be $\pi$, proving your assertion.

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Edit: sorry, there used to be a completely wrong solution here (I thought that a certain singular curve was a projective line). Now it is fixed.

There is another solution using algebraic geometry. Identify the ellipse with the projective line by sending the two points where the line through the foci meets the ellipse to $0$ and $\infty$. The map we get by starting from a point on the ellipse, getting the second intersection of the line through it and $F$ with the ellipse, then getting the second intersection of the line through that point and $F'$ with the ellipse is an invertible algebraic map sending $0$ to $0$ and $\infty$ to $\infty$, so it must have the form $x \mapsto cx$ for some constant $c$ (depending on the eccentricity). Thus, the billiard ball approaches the horizontal line at an exponential rate.

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You can figure out what $c$ is by looking at the action on the tangent spaces at $0$ and $\infty$. Let the distance between the focii be $2a$ and let $R$ be the major radius. Then going through the first focus, as a map from the tangent space at $0$ to the tangent space at $\infty$, is dilation by $(R+a)/(R-a)$. Going back from $\infty$ to $0$ is dilation by that factor again. So $c=(R+a)^2/(R-a)^2$. –  David Speyer Mar 21 '11 at 20:13