Intersection of Common Tangents to Two Ellipses We have three points in a plane $X_1$, $X_2$ and $X$ such that $XX_1 = 9$, $XX_2 = 7$ and $X_1X_2 = 8$. Ellipses $e_1$ and $e_2$ are drawn using $X$, $X_1$ and $X$, $X_2$ as foci and they pass through $X_2$ and $X_1$ respectively. The common tangents to the ellipses intersect in $Y$. It's asked to find $XY^2$.
I have found so far that $Y$ will lie on $X_1X_2$ extended. Using this we can try to find a messy solution using co-ordinate, however I feel there should be a geometry proof using properties of ellipses.
Can anyone please help with how to proceed?

 A: Take the reflections of $X$ in both the common tangents and mark them as $X'$ and $X''$.
Join $X_1$ and $X_2$ to $X'$ and $X''$. $X_1X' = X_1X'' = 15$ and $X_2X' = X_2X'' = 17$.
This can be shown using two properties of ellipse:

*

*Sum of distances of any point on ellipse to foci is a constant.

*A ray passing through one of the foci bounces back to the other focus after reflecting off the surface of the ellipse. (Can be proven with ease using property 1).

Now $\bigtriangleup X_1X_2X'$ and $\bigtriangleup X_1X_2X''$ are right angled using Pythagoras' theorem. This implies $X'$, $X_1$ and $X''$ are collinear and $X_1X_2$ is the perpendicular bisector of $X'X''$.
Also notice $Y$ has to be the circumcenter of the $\bigtriangleup{XX''X'}$ since $Y$ lies on the perpendicular bisectors of $XX''$ and $XX'$. $\implies X''Y = XY$
Imagine $X_2$ at $(0,0)$, $X_1$ at $(8,0)$. Can be shown that $X$ lies at $(2, \pm\sqrt{45})$.
Let $Y$ be at $(8+x, 0)$.
$X''Y^2 = x^2 + 225 = XY^2 = (x+6)^2 + 45 \implies x = 12 \implies XY^2 = 369$.

