11. As shown in the figure, the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its left focus at $F$. A line passing through point $F$ intersects the ellipse at points $A$ and $B$. When the line $AB$ passes through one of the vertices of the ellipse, its inclination angle is exactly $60^{\circ}$.
（1）Find the eccentricity of the ellipse;
（2）The midpoint of segment $AB$ is $G$, and the perpendicular bisector of $AB$ intersects the $x$-axis and $y$-axis at points $D$ and $E$, respectively. Let the area of $\triangle GDF$ be $S_{1}$, and the area of $\triangle OED$ (where $O$ is the origin) be $S_{2}$. Find the range of $\frac{S_{1}}{S_{2}}$.
Solution (1) According to the problem, when the line $AB$ passes through the vertex $(0, b)$ of the ellipse, its angle of inclination is $60^{\circ}$.
Let $F(-c, 0)$, then $\frac{b}{c}=\tan 60^{\circ}=\sqrt{3}$, substituting $b=\sqrt{3} c$ into $a^{2}=b^{2}+c^{2}$, we get $a=2 c$, so the eccentricity of the ellipse is
$$
e=\frac{c}{a}=\frac{1}{2}
$$
4 points
(2) From (1), the equation of the ellipse can be set as $\frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{3 c^{2}}=1$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. According to the problem, the line $AB$ cannot be perpendicular to the $x, y$ axes, so we set the equation of the line $AB$ as $y=k(x+c)$, substituting it into
$$
3 x^{2}+4 y^{2}=12 c^{2}
$$

we get
$$
\left(4 k^{2}+3\right) x^{2}+8 c k^{2} x+4 k^{2} c^{2}-12 c^{2}=0
$$

then $x_{1}+x_{2}=\frac{-8 c k^{2}}{4 k^{2}+3}, y_{1}+y_{2}=\frac{6 c k}{4 k^{2}+3}$, so $G\left(\frac{-4 c k^{2}}{4 k^{2}+3}, \frac{3 c k}{4 k^{2}+3}\right)$.
9 points
Since $GD \perp AB$, we have $\frac{\frac{3 c k}{4 k^{2}+3}}{\frac{-4 c k^{2}}{4 k^{2}+3}-x_{D}} \times k=-1, x_{D}=\frac{-c k^{2}}{4 k^{2}+3}$.
11 points
Since $\triangle GFD \sim \triangle OED$, we have
$$
\frac{S_{1}}{S_{2}}=\frac{|GD|^{2}}{|OD|^{2}}=\frac{\left(\frac{-4 c k^{2}}{4 k^{2}+3}-\frac{-c k^{2}}{4 k^{2}+3}\right)^{2}+\left(\frac{3 c k}{4 k^{2}+3}\right)^{2}}{\left(\frac{-c k^{2}}{4 k^{2}+3}\right)^{2}}=\frac{\left(3 c k^{2}\right)^{2}+(3 c k)^{2}}{\left(c k^{2}\right)^{2}}=\frac{9 c^{2} k^{4}+9 c^{2} k^{2}}{c^{2} k^{4}}=9+\frac{9}{k^{2}}>9
$$

Therefore, the range of $\frac{S_{1}}{S_{2}}$ is $(9,+\infty)$.
14 points