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. If you consider that set of ordered pairs (point on the ellipse, line passing through the point on the ellipse and at least one focus), you see that this set defines an algebraic curve $C$. The natural map from $C$ to Identify the ellipse (that we get by forgetting with the projective line ) is two to one at every point except for by sending the two points where the line through the foci hits meets the ellipse , so by the Riemann-Hurwitz formula $C$ has genus zero, i.e. it is isomorphic to the projective line.
You have defined a map $f$ from $C$ to itself, and this map can be defined algebraically 0$ and is clearly invertible, so as a $\infty$. The map we get by starting from the projective line to itself it is a fractional linear transformation. The map $f \circ f$ has two obvious fixed points where point on the ellipse, getting the second intersection of the line through the foci intersects it and $F$ with the ellipse, and if we fix an identification then getting the second intersection of the line through that point and $C$ F'$ with the projective line ellipse is an invertible algebraic map sending the two fixed points $0$ to $0$ and $\infty$, then we see that the map \infty$ to $f$ \infty$, so it must have the form $f(x) = c/x$ x \mapsto cx$ for some constant $c$ (depending on the eccentricity of the ellipse)eccentricity). Thusas we iterate $f$, $f^n(x)$ tends to $0$ and $\infty$ (in other words, the billiard ball tends to a approaches the horizontal line), and it does so line at an exponential rate.