11. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with its left focus at $F$. A tangent line is drawn through a point $A$ on the ellipse $C$, intersecting the $y$-axis at point $Q$, and $O$ is the origin. If $\angle Q F O=45^{\circ}, \angle Q F A=30^{\circ}$, then the eccentricity of the ellipse is $\qquad$.
11. $\frac{\sqrt{6}}{3}$.

As shown in Figure 5.
Let $A\left(x_{0}, y_{0}\right)$.
Then the equation of the tangent line at point $A$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
Let the tangent line intersect the $x$-axis at point $P$. Then $P\left(\frac{a^{2}}{x_{0}}, 0\right), Q\left(0, \frac{b^{2}}{y_{0}}\right)$.
Thus, $\overrightarrow{F Q} \cdot \overrightarrow{F O}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot(c, 0)=c^{2}$,
$$
\begin{array}{l}
\overrightarrow{F Q} \cdot \overrightarrow{F A}=\left(c, \frac{b^{2}}{y_{0}}\right) \cdot\left(x_{0}+c, y_{0}\right)=c x_{0}+a^{2} . \\
\text { Therefore, } \frac{\cos \angle Q F O}{\cos \angle Q F A}=\frac{\overrightarrow{F Q}}{\overrightarrow{F Q} \cdot \overrightarrow{F O}} \cdot \overrightarrow{F A} \cdot \frac{|\overrightarrow{F Q}||\overrightarrow{F A}|}{|\overrightarrow{F Q}||\overrightarrow{F O}|} \\
=\frac{c^{2}}{c x_{0}+a^{2}} \cdot \frac{e x_{0}+a}{c}=\frac{c}{a} \cdot \frac{e x_{0}+a}{e x_{0}+a}=e .
\end{array}
$$

Therefore, the eccentricity of the ellipse is
$$
e=\frac{\cos \angle Q F O}{\cos \angle Q F A}=\frac{\cos 45^{\circ}}{\cos 30^{\circ}}=\frac{\sqrt{6}}{3} .
$$