The hyperbola $3 x^{2}-y^{2}+24 x+36=0$ has a focus $F$ and a right directrix $l$. 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 $C$.
14. Transform the hyperbola equation to $\frac{(x-4)^{2}}{4} \cdots \frac{y^{2}}{12}=1$, where the coordinates of $F$ are $(8,0)$, and the equation of the right directrix $l$ is $x=5$.

Using $F$ as the pole and $F x$ as the polar axis, let $\mathrm{e}$ be the eccentricity of the ellipse $C$, then the equation of $C$ is $\rho = \frac{3 e}{1 \cdots \cos \theta}$. Let $\theta=\frac{\pi}{4}$ or $\frac{5 \pi}{4}$, we have $|F A|=\frac{3 \mathrm{e}}{1-\frac{\sqrt{2}}{2} \mathrm{e}}, |F B|=\frac{3 \mathrm{e}}{1+\frac{\sqrt{2}}{2} \mathrm{e}}$. 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 $|A I|^{2}=|F A|^{2}+|F I|^{2}-\sqrt{2}|F A| \cdot|F I|, |B I|^{2}=|F B|^{2}+|F I|^{2}+\sqrt{2}|F B| \cdot|F I|, |A B|^{2}=|F A|^{2}+|F B|^{2}+2|F A| \cdot|F B|$. According to the problem, we have: $|A I|^{2} + |B I|^{2} > |A B|^{2}$. Therefore, we should have $2 c^{2}+\sqrt{2}(|F B| \cdots|F A|) c > 2|F A| \cdot|F B|$, i.e., $2 c^{2}+$

Since $\frac{a^{2}}{c}-c=3$, we have $c=\frac{3 \mathrm{e}^{2}}{1-\mathrm{e}^{2}}$. Substituting this into the above equation, we get
$$
\frac{\mathrm{e}^{2}}{\left(1-\mathrm{e}^{2}\right)^{2}}-\frac{\mathrm{e}^{2}}{\left(1-\mathrm{e}^{2}\right)\left(1-\frac{1}{2} \mathrm{e}^{2}\right)}>\frac{1}{1-\frac{1}{2} \mathrm{e}^{2}} \text{. }
$$

Solving this, we get $0<\mathrm{e}<\sqrt{2-\sqrt{2}}$.