11. (20 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through point $F_{1}$ with an inclination angle $\alpha$ intersects the ellipse $C$ at points $A$ and $B$. If $\cos \alpha=\frac{5}{13}$, $5 \overrightarrow{F_{1} A}=8 \overrightarrow{B F_{1}}$, and the distance from point $F_{2}$ to the line $A B$ is $\frac{72}{13}$.
(1) Find the equation of the ellipse $C$;
(2) If $P$ and $Q$ are two points on the right directrix of the ellipse $C$, and $|P Q|=3$, find the equation of the trajectory of the incenter $M$ of $\triangle F_{1} P Q$.
11. (1) The distance from point $F_{2}$ to the line $A B$ is
$$
\begin{array}{l}
d=2 c \sin \alpha=\frac{24}{13} c=\frac{72}{13} \\
\Rightarrow c=3 .
\end{array}
$$

Let the projections of points $A$ and $B$ on the left directrix of the ellipse be $A_{1}$ and $B_{1}$, respectively.
Given $5\left|\overrightarrow{F_{1} A}\right|=8\left|\overrightarrow{B F_{1}}\right|$, set
$\left|A F_{1}\right|=8 t,\left|B F_{1}\right|=5 t(t>0)$.
Draw $B T \perp A A_{1}$ at point $T$. Then
$|T A|=|A B| \cos \alpha=13 t \cos \alpha=5 t$.
By $\frac{\left|A F_{1}\right|}{\left|A A_{1}\right|}=\frac{\left|B F_{1}\right|}{\left|B B_{1}\right|}=e$
$\Rightarrow\left|A A_{1}\right|-\left|B B_{1}\right|=\frac{\left|A F_{1}\right|-\left|B F_{1}\right|}{e}=\frac{3 t}{e}=5 t$
$\Rightarrow e=\frac{c}{a}=\frac{3}{5} \Rightarrow a=\frac{5}{3} c$
$\Rightarrow a=5$.
Thus, the equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$.
(2) Let the incenter $M(x, y)$.
Then the radius of $\odot M$ is $r=\frac{a^{2}}{c}-x=\frac{25}{3}-x$,
$S_{\triangle F_{1} P Q}=\frac{1}{2} \times 3 \times\left(c+\frac{a^{2}}{c}\right)=17$.
Thus, $\left|P F_{1}\right|+\left|Q F_{1}\right|=\frac{34}{\frac{25}{3}-x}-3$.
Draw $M K \perp F_{1} Q$ at point $K$, and connect $M F_{1}$
Hence, $\left|K F_{1}\right|^{2}=\left|M F_{1}\right|^{2}-r^{2}$
$$
\begin{array}{l}
=(x+3)^{2}+y^{2}-\left(\frac{25}{3}-x\right)^{2} . \\
\text { Also, }\left|P F_{1}\right|+\left|Q F_{1}\right|=2\left|K F_{1}\right|+3 \text {, so } \\
\left|K F_{1}\right|=\frac{1}{2}\left(\left|P F_{1}\right|+\left|Q F_{1}\right|-3\right) \\
=\frac{17}{\frac{25}{3}-x}-3 \text {. } \\
\text { Therefore, }\left(\frac{17}{\frac{25}{3}-x}-3\right)^{2}=(x+3)^{2}+y^{2}-\left(\frac{25}{3}-x\right)^{2} \\
\Rightarrow 81(3 x-8)^{2} \\
\quad=(25-3 x)^{2}\left(204 x+9 y^{2}-544\right) .
\end{array}
$$