20. Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line is drawn through its left focus $F_{1}$ intersecting the ellipse at points $\mathrm{A}, \mathrm{B}$. $\mathrm{D}(a, \theta)$ is a point to the right of $F_{1}$, and lines $\mathrm{AD}$ and $\mathrm{BD}$ intersect the left directrix of the ellipse at points $M, N$. If the circle with diameter $M N$ passes exactly through $F_{1}$, find the value of $a$.
Given: $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$; the equation of AB is $y=k(x+3)$ (where $k$ is the slope).
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. From $\left\{\begin{array}{l}y=k(x+3) \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array} \Rightarrow\left(16+25 k^{2}\right) x^{2}+150 k^{2} x+225 k^{2}-400=0\right.$ we get
$$
x_{1}+x_{2}=-\frac{150 k^{2}}{16+25 k^{2}}, x_{1} x_{2}=-\frac{225 k^{2}-400}{16+25 k^{2}} \Rightarrow y_{1} y_{2}=k^{2}\left(x_{1}+3\right)\left(x_{2}+3\right)=-\frac{256 k^{2}}{16+25 k^{2}}
$$
10 points
Let $M\left(-\frac{25}{3}, y_{3}\right), N\left(-\frac{25}{3}, y_{4}\right)$. Since M, A, D are collinear, $y_{3}=\frac{(3 a+25) y_{1}}{3\left(a-x_{1}\right)}$, similarly $y_{4}=\frac{(3 a+25) y_{2}}{3\left(a-x_{2}\right)}$.
Also, $\overrightarrow{F_{1} M}=\left(-\frac{16}{3}, y_{3}\right), \overrightarrow{F_{1} N}=\left(-\frac{16}{3}, y_{4}\right)$. Given that $\overrightarrow{F_{1} M} \perp \overrightarrow{F_{1} N} \Rightarrow \overrightarrow{F_{1} M} \bullet \overrightarrow{F_{1} N}=0$, we get $y_{3} y_{4}=-\frac{256}{9}$, and $y_{3} y_{4}=\frac{(3 a+25)^{2} y_{1} y_{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}$, i.e., $-\frac{256 k^{2}}{16+25 k^{2}} \cdot \frac{(3 a+25)^{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}=-\frac{256}{9}$, simplifying to $\left(1+k^{2}\right)\left(16 a^{2}-400\right)=0 \Rightarrow a= \pm 5$, and since $a>-3$, we have $a=5$
-17 points