 Let $E$ be an ellipse with foci $A$ and $B$. Suppose there exists a parabola $P$ such that
$\bullet$ $P$ passes through $A$ and $B$,
$\bullet$ the focus $F$ of $P$ lies on $E$,
$\bullet$ the orthocenter $H$ of $\vartriangle F AB$ lies on the directrix of $P$.
If the major and minor axes of $E$ have lengths $50$ and $14$, respectively, compute $AH^2 + BH^2$.
1. **Identify the given parameters and properties:**
   - The ellipse \( E \) has major and minor axes of lengths 50 and 14, respectively.
   - The foci of the ellipse are \( A \) and \( B \).
   - There exists a parabola \( P \) that passes through \( A \) and \( B \).
   - The focus \( F \) of the parabola \( P \) lies on the ellipse \( E \).
   - The orthocenter \( H \) of the triangle \( \triangle FAB \) lies on the directrix of the parabola \( P \).

2. **Calculate the distance between the foci \( A \) and \( B \) of the ellipse:**
   - The distance between the foci of an ellipse is given by \( 2c \), where \( c = \sqrt{a^2 - b^2} \).
   - Here, \( a = \frac{50}{2} = 25 \) and \( b = \frac{14}{2} = 7 \).
   \[
   c = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24
   \]
   - Therefore, the distance between the foci \( A \) and \( B \) is \( 2c = 48 \).

3. **Determine the properties of the parabola \( P \):**
   - The focus \( F \) of the parabola \( P \) lies on the ellipse \( E \).
   - The orthocenter \( H \) of \( \triangle FAB \) lies on the directrix of the parabola \( P \).

4. **Use the properties of the orthocenter and the directrix:**
   - The distance from \( A \) to the directrix of the parabola is equal to the distance from \( A \) to the focus \( F \) of the parabola.
   - Since \( F \) lies on the ellipse, the maximum distance \( AF \) can be is the semi-major axis of the ellipse, which is 25.

5. **Calculate the distance from \( A \) to the directrix:**
   - Let \( D \) be the projection of \( A \) onto the directrix.
   - Since \( AD = AF = 25 \), and \( H \) lies on the directrix, the distance \( DH \) is half the distance between \( A \) and \( B \) due to symmetry.
   \[
   DH = \frac{AB}{2} = \frac{48}{2} = 24
   \]

6. **Calculate \( AH \) using the Pythagorean theorem:**
   \[
   AH^2 = AD^2 + DH^2 = 25^2 + 24^2 = 625 + 576 = 1201
   \]

7. **Since \( AH = BH \), calculate \( AH^2 + BH^2 \):**
   \[
   AH^2 + BH^2 = 1201 + 1201 = 2402
   \]

The final answer is \(\boxed{2402}\)