11. Given that $A, B$ are the left and right vertices of the ellipse $C: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, and the line $l$ intersects the ellipse $C$ at points $M, N$. Let the slope of $A M$ be $k_{1}$, and the slope of $B N$ be $k_{2}$, with $k_{1}: k_{2}=1: 9$.
(1) Prove that the line $l$ passes through a fixed point;
(2) Let the area of $\triangle A M N$ be $S_{1}$, and the area of $\triangle B M N$ be $S_{2}$. Find the maximum value of $S_{1}-S_{2}$.
(1) Let the line $l: y=kx+m, M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, and combine $\left\{\begin{array}{l}y=kx+m, \\ 9x^{2}+25y^{2}=225\end{array}\right.$
$$
\Rightarrow\left(25k^{2}+9\right)x^{2}+50kmx+25m^{2}-225=0 .
$$

Then $k_{1}=\frac{y_{1}}{x_{1}+5}, k_{2}=\frac{y_{2}}{x_{2}-5}$, from the ellipse equation we get
$$
\frac{x_{1}^{2}}{25}+\frac{y_{1}^{2}}{9}=1 \Rightarrow \frac{y_{1}^{2}}{9}=\frac{\left(5+x_{1}\right)\left(5-x_{1}\right)}{25} \Rightarrow \frac{y_{1}}{x_{1}+5}=\frac{9\left(5-x_{1}\right)}{25y_{1}},
$$

Thus $k_{2}=9k_{1} \Rightarrow \frac{y_{2}}{x_{2}-5}=\frac{81}{25} \cdot \frac{5-x_{1}}{y_{1}} \Rightarrow \frac{y_{1}y_{2}}{\left(5-x_{1}\right)\left(x_{2}-5\right)}=\frac{k^{2}x_{1}x_{2}+km\left(x_{1}+x_{2}\right)+m^{2}}{-x_{1}x_{2}+5\left(x_{1}+x_{2}\right)-25}=\frac{81}{25}$
$$
\Rightarrow \frac{-25k^{2}+m^{2}}{-m^{2}-10km-25k^{2}}=9 \Rightarrow 20k^{2}+9km+m^{2}=0 \Rightarrow(4k+m)(5k+m)=0,
$$

Clearly, $5k+m=0 \Rightarrow$ the fixed point coordinates are $(5,0)$, which does not meet the problem's requirements;
So $4k+m=0 \Rightarrow$ the line $l$ passes through the fixed point $(4,0)$.
(2) Substitute $m=-4k$ into equation $(*)$ to get
$$
\begin{array}{l}
\left(25k^{2}+9\right)x^{2}-200k^{2}x+400k^{2}-225=0 . \\
\text { By } S_{1}=9S_{2} \Rightarrow S_{1}-S_{2}=8S_{2}=4\left|y_{1}-y_{2}\right| \\
=4\left|k \cdot \frac{\sqrt{200^{2}k^{4}-4\left(25k^{2}+9\right)\left(400k^{2}-225\right)}}{25k^{2}+9}\right| \\
=\frac{360|k| \sqrt{k^{2}+1}}{25k^{2}+9}=\frac{15 \cdot 2 \cdot 4|k| \sqrt{9k^{2}+9}}{25k^{2}+9} \leqslant \frac{15\left(16k^{2}+9k^{2}+9\right)}{25k^{2}+9}=15,
\end{array}
$$

Equality holds when $16k^{2}=9k^{2}+9 \Rightarrow k^{2}=\frac{9}{7}$. Therefore, the maximum value of $S_{1}-S_{2}$ is 15.