10. Since $2 a=\frac{a^{2}}{c}$, therefore, $c=\frac{1}{2} a, b^{2}=\frac{3}{4} a^{2}$.
Thus, the ellipse $C: 3 x^{2}+4 y^{2}=3 a^{2}$.
Let the line $l: x=m y+2 a$.
Combining the above two equations, we get
$$
\left(3 m^{2}+4\right) y^{2}+12 \text { may }+9 a^{2}=0 \text {. }
$$

According to the problem, the discriminant $\Delta>0$, i.e.,
$$
m^{2}-4>0 \text {. }
$$

Let the distances from points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ to the right directrix be $d_{1}$ and $d_{2}$, respectively. Then
$$
\begin{array}{l}
|A F|=\frac{1}{2} d_{1},|B F|=\frac{1}{2} d_{2} . \\
\text { Also, } d_{1}=\left|m y_{1}\right|, d_{2}=\left|m y_{2}\right|, \text { thus, } \\
\frac{1}{|A F|}+\frac{1}{|B F|}=\frac{1}{12} \\
\Rightarrow\left|\frac{1}{y_{1}}+\frac{1}{y_{2}}\right|=\frac{|m|}{24} .
\end{array}
$$

Substituting $y_{1}+y_{2}=\frac{-12 m a}{3 m^{2}+4}, y_{1} y_{2}=\frac{9 a^{2}}{3 m^{2}+4}$ into equation (1), we get
$$
\frac{|12 m a|}{9 a^{2}}=\frac{|m|}{24} \Rightarrow a=32 .
$$

Then $S_{\triangle A B F}=\frac{1}{2} \times \frac{3 a}{2}\left|y_{1}-y_{2}\right|$
$$
\begin{array}{l}
=\frac{9 a^{2}}{2} \cdot \frac{\sqrt{m^{2}-4}}{3 m^{2}+4} . \\
\text { Let } t=\sqrt{m^{2}-4}(t>0) .
\end{array}
$$

Then equation (2) can be transformed into
$$
\begin{array}{l}
S_{\triangle A B F}=\frac{9 a^{2}}{2} \cdot \frac{t}{3 t^{2}+16} \\
\leqslant \frac{9 a^{2}}{2} \cdot \frac{t}{2 \times 4 \times \sqrt{3} t}=192 \sqrt{3} .
\end{array}
$$

When and only when $t^{2}=\frac{16}{3}$, i.e., $m= \pm \frac{2 \sqrt{21}}{3}$,
$$
\left(S_{\triangle A B F}\right)_{\max }=192 \sqrt{3} .
$$
10. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$, $F$ is the right focus of the ellipse $C$. A line $l$ is drawn through the intersection of the right directrix $x=2a$ and the $x$-axis, intersecting the ellipse $C$ at points $A$ and $B$. The arithmetic mean of $\frac{1}{|A F|}$ and $\frac{1}{|B F|}$ is $\frac{1}{24}$. Find the maximum area of $\triangle A B F$.