, 2022

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 length formula. Thus $A H^{2} + B H^{2} = 2\left(25^{2} + 24^{2}\right) = 2402$.