11. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, a line $l$ passing through the left focus $F(-c, 0)(c>0)$ of the ellipse intersects the ellipse at points $A, B$. The perpendicular bisector of segment $A B$ intersects the ellipse at points $C, D$. If $A C \perp A D$, find the equation of line $l$.
According to the projection theorem, we have $|A M|^{2}=|C M| \cdot|M D|$
$$
\Rightarrow|A M| \cdot|M B|=|C M| \cdot|M D| \Rightarrow A, B, C, D
$$

are concyclic. Thus, let $M\left(x_{0}, y_{0}\right)$,
the line $A B:\left\{\begin{array}{l}x=x_{0}+t \cos \alpha, \\ y=y_{0}+t \sin \alpha,\end{array}\right.$
substituting into the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ yields
$$
\begin{array}{l}
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2\left(b^{2} x_{0} \cos \alpha+a^{2} y_{0} \sin \alpha\right) t+b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}=0 \\
\Rightarrow|A M| \cdot|M B|=\left|t_{1} t_{2}\right|=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2}+c^{2} \sin ^{2} \alpha},
\end{array}
$$

Similarly, for $C D:\left\{\begin{array}{l}x=x_{0}+t \cos \beta, \\ y=y_{0}+t \sin \beta,\end{array} \Rightarrow|C M| \cdot|M D|=\frac{\left|b^{2} x_{0}^{2}+a^{2} y_{0}^{2}-a^{2} b^{2}\right|}{b^{2}+c^{2} \sin ^{2} \beta}\right.$,
thus $\sin \alpha=\sin \beta \Rightarrow \alpha+\beta=180^{\circ} \Rightarrow k_{A B}+k_{C D}=0$, and $k_{A B} \cdot k_{C D}=-1 \Rightarrow k_{A B}= \pm 1$, so the equation of the line $l$ is $y= \pm x+c$.