11. (20 points) Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ and three fixed points $A(-1,0), B(1,0), C(3,0)$, a line passing through $C$ intersects the ellipse at points $P, Q$ (point $Q$ is to the left of point $P$), and the lines $A P$ and $B Q$ intersect at point $R$. Find the locus of point $R$.
11. Let $R(x, y)$ be any point on the desired trajectory. Let $Q P=t P C$. By Menelaus' theorem, we have
$$
1=\frac{B R}{R Q} \cdot \frac{Q P}{P C} \cdot \frac{C A}{A B}=\frac{B R}{R Q} \cdot 2 t \Rightarrow \frac{R Q}{B R}=2 t \Rightarrow \frac{Q B}{R B}=1+\frac{R Q}{B R}=1+2 t .
$$

Thus, the coordinates of point $Q$ are $((2 t+1)(x-1)+1,(2 t+1) y)$, and $P\left(\frac{2 t+1}{t+1}(x+1)-1, \frac{2 t+1}{t+1} y\right)$. Since $P, Q$ are both on the ellipse, we have
$$
\begin{array}{l}
((2 t+1)(x+1)-(t+1))^{2}+4 \cdot((2 t+1) y)^{2}=4(t+1)^{2} \\
((2 t+1)(x-1)+1)^{2}+4 \cdot((2 t+1) y)^{2}=4
\end{array}
$$

Subtracting the two equations, we get
$$
3 t \cdot((2 t+1) \cdot 2 x-t)=4 t \cdot(t+2) \text {. }
$$

Since $t>0$, we have
$$
12 x(2 t+1)=7(2 t+1)+9 \Rightarrow 2 t+1=\frac{9}{12 x-7} .
$$

Substituting back into equation (2), we get
$$
\left(\frac{9(x-1)}{12 x-7}+1\right)^{2}+4 \cdot\left(\frac{9 y}{12 x-7}\right)^{2}=4
$$

which simplifies to $45 x^{2}-108 y^{2}=20$. This is the equation of the trajectory of point $R$. Considering the graph, we can only take the right branch of the hyperbola and the part inside the ellipse, excluding the vertex (since when $P, Q$ are the vertices of the ellipse, the lines $A P, B Q$ coincide and cannot determine the intersection point $R$). Therefore, the trajectory of the intersection point $R$ is $45 x^{2}-108 y^{2}=20, x \in\left(\frac{2}{3}, \frac{4}{3}\right)$.