11. (15 points) The right focus of the hyperbola $3 x^{2}-y^{2}+24 x+36=0$ is $F$, and the right directrix is $l$. The ellipse $C$ has $F$ and $l$ as its corresponding focus and directrix. A line parallel to $y=x$ is drawn through $F$, intersecting the ellipse $C$ at points $A$ and $B$. It is known that the center of $C$ is inside the circle with $A B$ as its diameter. Find the range of the eccentricity of the ellipse $C$.
11. Solution: Transform the hyperbola equation to $\frac{(x+4)^{2}}{4}-\frac{y^{2}}{12}=1$, then the coordinates of $F$ are $(0,0)$, and the equation of the right directrix $l$ is $x=-3$. Establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis, and let $e$ be the eccentricity of the ellipse $C$. Then the equation of $C$ is $\rho=\frac{3 e}{1-e \cos \theta}$. Let $\theta=\frac{\pi}{4}$ or $\frac{5 \pi}{4}$, we have $|F A|=\frac{3 e}{1-\frac{\sqrt{2}}{2} e}$ and $|F B|=\frac{3 e}{1+\frac{\sqrt{2}}{2} e}$. On the other hand, let the center of $C$ be $I$, then $|F I|=c$, and $\frac{a^{2}}{c}-c=3$ (where $a, b, c$ are the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse $C$, respectively). Using the cosine rule, we know that $|A I|^{2}=|F A|^{2}+|F I|^{2}-\sqrt{2}|F A| \cdot|F I|$, $|B I|^{2}=|F B|^{2}+|F I|^{2}+\sqrt{2}|F B| \cdot|F I|$, and $|A B|^{2}=|F A|^{2}+|F B|^{2}+2|F A| \cdot|F B|$. According to the problem, $|A I|^{2}+|B I|^{2}>|A B|^{2}$. Therefore, we should have $2 c^{2}+\sqrt{2}(|F B|-|F A|) c>2|F A| \cdot|F B|$, which is
$2 c^{2}+\sqrt{2} c\left(\frac{3 e}{1+\frac{\sqrt{2}}{2} e}-\frac{3 e}{1-\frac{\sqrt{2}}{2} e}\right)>\frac{18 e^{2}}{1-\frac{1}{2} e^{2}}$. Since $\frac{a^{2}}{c}-c=3$, we have $c=\frac{3 e^{2}}{1-e^{2}}$.

Substituting into the above equation, we get $\frac{e^{2}}{\left(1-e^{2}\right)^{2}}-\frac{e^{2}}{\left(1-e^{2}\right)\left(1-\frac{1}{2} e^{2}\right)}>\frac{1}{1-\frac{1}{2} e^{2}}$. Solving this, we get $0<e<\sqrt{2-\sqrt{2}}$.