11. (20 points) For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, do there exist two fixed points $A$ and $B$ such that for any point $M$ on the ellipse, if the slopes $k_{1}$ and $k_{2}$ of $AM$ and $BM$ exist, then $k_{1} k_{2}$ is a constant? If they exist, find all such pairs of points; if not, explain the reason.
11. Suppose there exist fixed points $A, B$, and let
$$
\begin{array}{l}
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M(x, y). \\
\text { Then } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \text { and } \\
k_{1} k_{2}=\frac{y-y_{1}}{x-x_{1}} \cdot \frac{y-y_{2}}{x-x_{2}} \\
=\frac{y^{2}-\left(y_{1}+y_{2}\right) y+y_{1} y_{2}}{x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}}=t \text { (constant). } \\
\text { Hence } y^{2}-\left(y_{1}+y_{2}\right) y+y_{1} y_{2} \\
=t\left(x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}\right)
\end{array}
$$

This holds for any point $(x, y)$ on the ellipse.
By the symmetry of the ellipse, the above equation also holds for the point $(-x, -y)$, i.e.,
$$
\begin{array}{l}
y^{2}+\left(y_{1}+y_{2}\right) y+y_{1} y_{2} \\
=t\left(x^{2}+\left(x_{1}+x_{2}\right) x+x_{1} x_{2}\right).
\end{array}
$$

Thus, $\left(y_{1}+y_{2}\right) y=t\left(x_{1}+x_{2}\right) x$ holds for any point $(x, y)$ on the ellipse.
Therefore, $x_{1}+x_{2}=0, y_{1}+y_{2}=0$.
Hence, from equation (1) we have
$$
y^{2}-y_{1}^{2}=t\left(x^{2}-x_{1}^{2}\right)=t a^{2}-t \frac{a^{2} y^{2}}{b^{2}}-t x_{1}^{2}
$$

holds for any point $(x, y)$ on the ellipse.
Therefore, $t=-\frac{b^{2}}{a^{2}}, \frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$, which means points $A, B$ are on the ellipse and symmetric about the origin. Clearly, $k_{1} k_{2}$ is a constant.
In summary, there exist such fixed points $A, B$, both on the ellipse and symmetric about the origin, such that $k_{1} k_{2}=-\frac{b^{2}}{a^{2}}$ holds.