(20) Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line through its left focus $F_{1}$ intersects the ellipse at points $A$ and $B$. Point $D(a, 0)$ is to the right of $F_{1}$. Lines $A D$ and $B D$ intersect the left directrix of the ellipse at points $M$ and $N$, respectively. If the circle with diameter $M N$ passes through $F_{1}$, find the value of $a$.
(20) $F_{1}(-3,0)$, the left directrix equation is $x=-\frac{25}{3}$; $A B$ equation is $y=k(x+3)$ ( $k$ is the slope).
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
$$
\left\{\begin{array}{l}
y=k(x+3), \\
\frac{x^{2}}{25}+\frac{y^{2}}{16}=1,
\end{array} \Rightarrow\left(16+25 k^{2}\right) x^{2}+150 k^{2} x+225 k^{2}-400=0,\right.
$$

we get $x_{1}+x_{2}=-\frac{150 k^{2}}{16+25 k^{2}}, x_{1} x_{2}=-\frac{225 k^{2}-400}{16+25 k^{2}}$,
$$
y_{1} y_{2}=k^{2}\left(x_{1}+3\right)\left(x_{2}+3\right)=-\frac{256 k^{2}}{16+25 k^{2}} .
$$

Let $M\left(-\frac{25}{3}, y_{3}\right), N\left(-\frac{25}{3}, y_{4}\right)$. From the collinearity of $M, A, D$ we get
$$
y_{3}=\frac{(3 a+25) y_{1}}{3\left(a-x_{1}\right)},
$$

Similarly,
$$
y_{4}=\frac{(3 a+25) y_{2}}{3\left(a-x_{2}\right)} .
$$

Also, $\overrightarrow{F_{1} M}=\left(-\frac{16}{3}, y_{3}\right), \overrightarrow{F_{1} N}=\left(-\frac{16}{3}, y_{4}\right)$, and by the given condition
$$
\overrightarrow{F_{1} M} \perp \overrightarrow{F_{1} N} \Rightarrow \overrightarrow{F_{1} M} \cdot \overrightarrow{F_{1} N}=0,
$$

we get $y_{3} y_{4}=-\frac{256}{9}$, and
$$
y_{3} y_{4}=\frac{(3 a+25)^{2} y_{1} y_{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)},
$$

which means
$$
-\frac{256 k^{2}}{16+25 k^{2}} \cdot \frac{(3 a+25)^{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}=-\frac{256}{9},
$$

simplifying to
$$
\left(1+k^{2}\right)\left(16 a^{2}-400\right)=0 \Rightarrow a= \pm 5,
$$

and since $a>-3$, we have $a=5$.