20. It is known that $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$, and $l_{A B}: y=k(x+3)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k(x+3), \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array}\right.$
$$\Rightarrow\left(16+25 k^{2}\right) x^{2}+150 k^{2} x+225 k^{2}-400=0 .$$

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

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
$$y_{3}=\frac{(3 a+25) y_{1}}{3\left(a-x_{1}\right)} \text {. }$$

Similarly, $y_{4}=\frac{(3 a+25) y_{2}}{3\left(a-x_{2}\right)}$.
$$\text { Also, } \overrightarrow{F_{1} M}=\left(-\frac{16}{3}, y_{3}\right), \overrightarrow{F_{1} N}=\left(-\frac{16}{3}, y_{4}\right) \text {. }$$

From the given information,
$$\begin{array}{l} \overrightarrow{F_{1} M} \perp \overrightarrow{F_{1} N} \Rightarrow \overrightarrow{F_{1} M} \cdot \overrightarrow{F_{1} N}=0 \\ \Rightarrow y_{3} y_{4}=-\frac{256}{9} \text {. } \\ \end{array}$$
$$\begin{array}{l} \text { And } y_{3} y_{4}=\frac{(3 a+25)^{2} y_{1} y_{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)} \\ \Rightarrow-\frac{256 k^{2}}{16+25 k^{2}} \cdot \frac{(3 a+25)^{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}=-\frac{256}{9} \\ \Rightarrow\left(1+k^{2}\right)\left(16 a^{2}-400\right)=0 \end{array}$$
$\Rightarrow a= \pm 5$ (negative value is discarded).