Alhazen's Billiard Problem in an Ellipse This was a question that was for some reason removed by the OP. I'm a prospective sophomore intrigued by the Alhazen's billiard problem. Could someone post the answer to the original question here?
I was working on a project for school, and was wondering how to go about it in an unconventional manner. I thought of solving Alhazen's Billiard Problem, that is, deriving a general equation for it, for an ellipse. This problem is usually done for a circle, and so far, I've tried to use the method provided in Heinrich Dorrie's '100 Great Problems of Elementary Mathematics.' This problem concerns reflections, and is solved using tan ratios of the angles that the incident and reflected ray strike with. Here is a diagram for your reference (of Dorrie's method that I've tried to apply here).

When I equate the tan ratios of alpha and beta using the other angle sums mentioned, the substitution gets messy due to the ellipse equation. Any outlook or thought would be appreciated. Thank you!
 A: Assume that $A$ and $B$ are two points inside a circular billiard $\Gamma$ with centre $O$.
We want a point $P$ on the boundary of $\Gamma$ such that the shot from $A$ to $P$ goes through $B$:

so we want a point $P\in\partial\Gamma$ such that $OP$ bisects $\widehat{APB}$. We have (at least) two ways for approaching the question:


*

*We may consider a variable $P(\theta)$ on $\partial\Gamma$ and compute $Q(\theta) = OP\cap AB$. By the bisector theorem, $OP$ bisects $\widehat{APB}$ iff $\frac{AQ}{QB}=\frac{AP}{PB}$;

*We may consider the ellipse through $P$ with foci at $A,B$. $P$ is a solution iff such ellipse is tangent to $\Gamma$ at $P$, hence we may consider the pencil of ellipses with foci at $A$ and $B$ and compute the parameter for which a member of such pencil is tangent to $\Gamma$ by imposing that a discriminant is vanishing.


In both cases, the solution is given by a root of a third degree polynomial, so the problem, in general, is not solvable through straightedge and compass.
A: Could you apply our recent results? and we are considering the applications to Geometry.
The division by zero is uniquely and reasonably determined as 1/0=0/0=z/0=0 in the natural extensions of fractions. We have to change our basic ideas for our space and world
Division by Zero z/0 = 0 in Euclidean Spaces
Hiroshi Michiwaki, Hiroshi Okumura and Saburou Saitoh
International Journal of Mathematics and Computation Vol. 28(2017); Issue  1, 2017), 1 
-16.　 
http://www.scirp.org/journal/alamt　 　http://dx.doi.org/10.4236/alamt.2016.62007
http://www.ijapm.org/show-63-504-1.html
http://www.diogenes.bg/ijam/contents/2014-27-2/9/9.pdf
http://okmr.yamatoblog.net/division%20by%20zero/announcement%20326-%20the%20divi
http://okmr.yamatoblog.net/
Relations of 0 and infinity
Hiroshi Okumura, Saburou Saitoh　and Tsutomu Matsuura：
http://www.e-jikei.org/…/Camera%20ready%20manuscript_JTSS_A…
https://sites.google.com/site/sandrapinelas/icddea-2017
