11. In rectangle $A B C D$, it is known that $A B=2, A D<\sqrt{2}$, and an ellipse $K$ is constructed with side $A B$ as the major axis such that the length of the minor axis of ellipse $K$ is $\sqrt{2}|A D|$. Take a point $P$ on ellipse $K$ different from the endpoints, and connect $P C$ and $P D$, intersecting $A B$ at $M$ and $L$ respectively. Find the value of $A M^{2}+L B^{2}$.
11. Solution: Taking the midpoint $O$ of $AB$ as the origin, establish a rectangular coordinate system, and let $|AD|=u$. Then the equation of the ellipse is $x^{2}+\frac{2 y^{2}}{u^{2}}=1$. Let point $P\left(\cos \theta, \frac{u}{\sqrt{2}} \sin \theta\right)$. The equation of line $DP$ is $y-u=\frac{u \sin \theta-\sqrt{2} u}{\sqrt{2}(1+\cos \theta)}(x+1)$. Thus, the x-coordinate of $L$ is $x_{1}=\frac{\sqrt{2}(1+\cos \theta)}{\sqrt{2}-\sin \theta}-1$,
Similarly, $x_{M}=\frac{\sqrt{2}(\cos \theta-1)}{\sqrt{2}-\sin \theta}+1$,
Therefore, $A M^{2}+L B^{2}=\left[\frac{\sqrt{2}(\cos \theta+1)}{\sqrt{2}-\sin \theta}-2\right]^{2}+\left[\frac{\sqrt{2}(\cos \theta-1)}{\sqrt{2}-\sin \theta}+2\right]^{2}$
$$
\begin{array}{l}
=\frac{4\left(2-\sin ^{2} \theta\right)-8 \sqrt{2}(\sqrt{2}-\sin \theta)}{(\sqrt{2}-\sin \theta)^{2}}+8=\frac{-8+8 \sqrt{2} \sin \theta-4 \sin ^{2} \theta}{(\sqrt{2}-\sin \theta)^{2}}+8 \\
=\frac{-4(\sqrt{2}-\sin \theta)^{2}}{(\sqrt{2}-\sin \theta)^{2}}+8=4 .
\end{array}
$$