10. (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 $\theta$ intersects the ellipse $C$ at points $A$ and $B$. If $\cos \theta=\frac{3}{5}$, $3 \overrightarrow{F A_{1}}=5 \overrightarrow{B F_{1}}$, and $S_{\triangle A B F_{2}}=\frac{3808}{45}$.
(1) Find the equation of the ellipse $C$;
(2) If $P$ and $Q$ are two moving points on the right directrix of the ellipse $C$, and $|P Q|=10$, find the equation of the trajectory of the incenter $M$ of $\triangle F_{1} P Q$.


10. (1) Let the projections of points $A$ and $B$ on the left directrix of the ellipse be points $A_{1}$ and $B_{1}$, respectively.
Given $3 \overrightarrow{F_{1} A}=5 \overrightarrow{B F_{1}}$, let $\left|A F_{1}\right|=5 t$ and $\left|B F_{1}\right|=3 t$ (where $t>0$).
Draw $B T \perp A A_{1}$ at point $T$.
Then $|T A|=|A B| \cos \theta=8 t \cos \theta=\frac{24}{5} t$.
By $\frac{\left|A F_{1}\right|}{\left|A A_{1}\right|}=\frac{\left|B F_{1}\right|}{\left|B B_{1}\right|}=e \Rightarrow\left\{\begin{array}{l}\left|A A_{1}\right|=\frac{5 t}{e} \\ \left|B B_{1}\right|=\frac{3 t}{e}\end{array}\right.$,
$\Rightarrow|T A|=\left|A A_{1}\right|-\left|B B_{1}\right|=\frac{2 t}{e}=\frac{24}{5} t$
$\Rightarrow e=\frac{c}{a}=\frac{5}{12} \Rightarrow a=\frac{12}{5} c$.
Also, $S_{\triangle A B F_{2}}=\frac{1}{2}|A B| \cdot 2 c \sin \theta=\frac{32}{5} t c$.
Using the cosine rule, we get
$$
\begin{array}{l}
\cos \theta=\frac{3}{5}=\frac{\left|A F_{1}\right|^{2}+\left|F_{1} F_{2}\right|^{2}-\left|A F_{2}\right|^{2}}{2\left|A F_{1}\right|\left|F_{1} F_{2}\right|} \\
=\frac{25 t^{2}+4 c^{2}-(2 a-5 t)^{2}}{2 \times 5 t \times 2 c} \\
\Rightarrow 3 t c=c^{2}-a^{2}+5 a t .
\end{array}
$$

Substituting $a=\frac{12}{5} c$ into the equation (1) gives $t=\frac{119}{225} c$.
Then $S_{\triangle A B F_{2}}=\frac{32}{5} \times \frac{119}{225} c^{2}=\frac{3808}{45}$
$\Rightarrow c^{2}=25, a^{2}=\frac{144}{25} c^{2}=144$
$$
\Rightarrow b^{2}=a^{2}-c^{2}=119 \text {. }
$$

Thus, the equation of the ellipse $C$ is $\frac{x^{2}}{144}+\frac{y^{2}}{119}=1$.
(2) Let point $M(x, y)$.

Then the radius of the incircle $\odot M$ is $r=\frac{a^{2}}{c}-x=\frac{144}{5}-x$,
$$
\begin{array}{l}
S_{\triangle F_{1} P Q}=\frac{1}{2} \times 10\left(c+\frac{a^{2}}{c}\right)=169 . \\
\text { Also, } S_{\triangle F_{1} P Q}=\frac{1}{2} r\left(\left|P F_{1}\right|+\left|Q F_{1}\right|+10\right) \\
=\frac{1}{2}\left(\frac{144}{5}-x\right)\left(\left|P F_{1}\right|+\left|Q F_{1}\right|+10\right)=169 \\
\Rightarrow\left|P F_{1}\right|+\left|Q F_{1}\right|=\frac{338}{\frac{144}{5}-x}-10 .
\end{array}
$$

Draw $M K \perp F_{1} Q$ at point $K$, and connect $M F_{1}$. Then
$$
\begin{array}{l}
\left|K F_{1}\right|^{2}=\left|M F_{1}\right|^{2}-r^{2}=(x+5)^{2}+y^{2}-\left(\frac{144}{5}-x\right)^{2} \\
=\frac{338}{5} x+y^{2}-\frac{169 \times 119}{25} . \\
\text { Also, }\left|P F_{1}\right|+\left|Q F_{1}\right|=2\left|K F_{1}\right|+10 \\
\Rightarrow 2\left|K F_{1}\right|=\frac{338}{\frac{144}{5}-x}-20 \\
\Rightarrow\left|K F_{1}\right|=\frac{119}{\frac{144}{5}-x}-10 .
\end{array}
$$

Therefore, the equation of the locus of point $M$ is
$$
\left(\frac{x-169}{\frac{144}{5}-x}\right)^{2}=\frac{338}{5} x+y^{2}-\frac{169 \times 119}{25} .
$$