8. Let $\mathcal{E}$ be an ellipse with foci $A$ and $B$. Suppose there exists a parabola $\mathcal{P}$ such that
- $\mathcal{P}$ passes through $A$ and $B$,
- the focus $F$ of $\mathcal{P}$ lies on $\mathcal{E}$,
- the orthocenter $H$ of $\triangle F A B$ lies on the directrix of $\mathcal{P}$.
If the major and minor axes of $\mathcal{E}$ have lengths 50 and 14 , respectively, compute $A H^{2}+B H^{2}$.
Answer: 2402

Solution: Let $D$ and $E$ be the projections of $A$ and $B$ onto the directrix of $\mathcal{P}$, respectively. Also, let $\omega_{A}$ be the circle centered at $A$ with radius $A D=A F$, and define $\omega_{B}$ similarly.
If $M$ is the midpoint of $\overline{D E}$, then $M$ lies on the radical axis of $\omega_{A}$ and $\omega_{B}$ since $M D^{2}=M E^{2}$. Since $F$ lies on both $\omega_{A}$ and $\omega_{B}$, it follows that $M F$ is the radical axis of the two circles. Moreover, $M F \perp A B$, so we must have $M=H$.
Let $N$ be the midpoint of $\overline{A B}$. We compute that $A D+B E=A F+F B=50$, so $H N=\frac{1}{2}(A D+B E)=$ 25. Since $A B=2 \sqrt{25^{2}-7^{2}}=48$, we have
$$
\begin{aligned}
25^{2}=H N^{2} & =\frac{1}{2}\left(A H^{2}+B H^{2}\right)-\frac{1}{4} A B^{2} \\
& =\frac{1}{2}\left(A H^{2}+B H^{2}\right)-24^{2}
\end{aligned}
$$
by the median lengtb formula. Thus $A H^{2}+B H^{2}=2\left(25^{2}+24^{2}\right)=2402$.