11. (15 points) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, a line $l$ passing through the left focus $F(-c, 0)$ of the ellipse intersects the ellipse at points $A$ and $B$. The perpendicular bisector of segment $AB$ intersects the ellipse at points $C$ and $D$. If $AC \perp AD$, find the equation of line $l$.

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II. 11. Let the line $l: y=k(x+c)$. Then $l_{C D}: x=-k y+b$.
Since $A C \perp A D$, and $C D$ is the perpendicular bisector of segment $A B$, it follows that $B C \perp B D$.
Therefore, points $A, B, C, D$ are concyclic.
Thus, the equation of the curve passing through points $A, B, C, D$ is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1+\lambda(k x-y-k c)(x+k y-b)=0 \text {. }
$$

Since points $A, B, C, D$ are concyclic, there is no $x y$ term in the equation.
Hence $\lambda\left(k^{2}-1\right)=0 \Rightarrow k= \pm 1$
$\Rightarrow$ the line $l: y= \pm(x+c)$.