11. Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ pass through the point $M(\sqrt{2}, 1)$, and have the left focus at $F_{1}(-\sqrt{2}, 0)$.
(1) Find the equation of the ellipse $C$;
(2) When a moving line $l$ passing through the point $P(4,1)$ intersects the ellipse $C$ at two distinct points $A, B$, take a point $Q$ on the line segment $A B$ such that $|\overrightarrow{A P}| \cdot|\overrightarrow{Q B}|=|\overrightarrow{A Q}| \cdot|\overrightarrow{P B}|$, find the equation of the trajectory of point $Q$.
11. Solution: (1) From the given, we have $c=\sqrt{2}, F_{2}(\sqrt{2}, 0)$, thus $a=\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{2}=\frac{3+1}{2}=2, b^{2}=a^{2}-c^{2}=2$, the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$. (2) Let $l: y-1=k(x-4), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), Q(x, y)$. From $\left\{\begin{array}{l}y=k x+1-4 k \\ x^{2}+2 y^{2}=4\end{array}\right.$, eliminating $y$, we get $x^{2}+2(k x+1-4 k)^{2}=4$, rearranging gives $\left(2 k^{2}+1\right) x^{2}+4 k(1-4 k) x+2(1-4 k)^{2}-4=0$. By Vieta's formulas, we have $x_{1}+x_{2}=\frac{4 k(4 k-1)}{2 k^{2}+1}, x_{1} x_{2}=\frac{2(4 k-1)^{2}-4}{2 k^{2}+1}$. Since $|\overrightarrow{A P}| \cdot|\overrightarrow{Q B}|=|\overrightarrow{A Q}| \cdot|\overrightarrow{P B}|$, we have $\sqrt{1+k^{2}}$ $\left|x_{1}-4\right| \cdot \sqrt{1+k^{2}}\left|x-x_{2}\right|=\sqrt{1+k^{2}}\left|x-x_{1}\right| \cdot \sqrt{1+k^{2}}\left|x_{2}-4\right|$, which simplifies to $\left|\left(x_{1}-4\right)\left(x-x_{2}\right)\right|=$ $\left|\left(x-x_{1}\right)\left(x_{2}-4\right)\right|$. Since $\left(x_{1}-4\right)\left(x-x_{2}\right)$ and $\left(x-x_{1}\right)\left(x_{2}-4\right)$ have opposite signs, $\left(x_{1}-4\right)\left(x-x_{2}\right)+\left(x_{2}-4\right)$ $\left(x-x_{1}\right)=0$, which simplifies to $\left(x_{1}+x_{2}\right)(x+4)-2 x_{1} x_{2}-8 x=0$, thus $\quad(x+4) \frac{k(4 k-1)}{2 k^{2}+1}-\frac{(4 k-1)^{2}-2}{2 k^{2}+1}-2 x=0$, which simplifies to $k(x-4)+2 x-1=0$. Substituting $k=\frac{y-1}{x-4}$, we rearrange to get $2 x+y-2=0$, where $\frac{8-2 \sqrt{7}}{9}<x<\frac{8+2 \sqrt{7}}{9}$.