The Prisoner's Dilemma

Robert Noll (PMP), Randy Schwartz (Schoolcraft College), and the Eagle Problem Solvers of Georgia Southern University submitted solutions.

The key observation is that if the elliptical track ever lies outside of the circle of radius two centered at (1, 0), then it will have to intersect that circle again for some $x$ value greater than -1. So we solve for the point of intersection of that circle and the ellipse that is stretched by a factor of $a$ in the $y$-direction, and determine for which $a$ there is an intersection with $x$-coordinate greater than -1.

The equation of the circle is $(x-1{)}^2+y^2=2^2$, which simplifies to $y^2=3+2x-x^2$. The equation of the ellipse is $x^2+y^2/a^2=1$, which simplifies to $y^2=a^2-a^{2}x$. Equating the two expressions for $y^2$ to find the points (or possibly point) of intersection yields $(a^2-1)x^2+2x+(3-a^2)=0$, with right-hand root

$$\begin{array}{c}\frac{-2+\sqrt{4-4(a^2-1)(3-a^2)}}{2(a^2-1)}=\frac{-2+\sqrt{4a^4-16a^2+16}}{2(a^2-1)}\hfill \\ \hfill =\frac{-2+2(a^2-2)}{2(a^2-1)}=\frac{a^2-3}{a^2-1}.\end{array}$$

Solving for the condition that this root is greater than -1 yields the condition that , which simplifies to . Thus, we conclude that the largest vertical stretch for which (-1,0) is the farthest point on the ellipse from (0, 1) occurs for $a=\sqrt{2}$. This stretch corresponds to an eccentricity of $e=\sqrt{1/2}$.