11. (20 points) If $N(m, 0)$ is a point on the internal axis of symmetry (the $x$-axis) of the quadratic curve $\Gamma$:
$$
A x^{2}+B y^{2}+D x+F=0
$$
$\left(A^{2}+B^{2} \neq 0,2 A m+D \neq 0\right)$, and a moving line through $N$ intersects the quadratic curve $\Gamma$ at points $P$ and $Q$. Let $M$ be a point on the external axis of symmetry (the $x$-axis) of the quadratic curve $\Gamma$. Then the necessary and sufficient condition for $\angle P M N=\angle Q M N$ is that the point $M\left(-\frac{D M+2 F}{2 A m+D}, 0\right)$.
11. Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)\left(x_{1}<m<x_{2}\right)$,
$$
\begin{array}{l}
M\left(x_{0}, 0\right), l_{P Q}: x=k y+m . \\
\quad \text { From }\left\{\begin{array}{l}
x=k y+m, \\
A x^{2}+B y+D x+F=0,
\end{array} \text { eliminate } x\right. \text { to get } \\
\quad\left(A k^{2}+B\right) y^{2}+(2 A m+D) k y+A m^{2}+D m+F=0 .
\end{array}
$$

From $\left\{\begin{array}{l}x=k y+m, \\ A x^{2}+B y+D x+F=0,\end{array}\right.$ eliminate $x$ to get

then $y_{1} y_{2}=\frac{A m^{2}+D m+F}{A k^{2}+B}$,
$$
y_{1}+y_{2}=-\frac{(2 A m+D) k}{A k^{2}+B} \text {. }
$$

From $\angle P M N=\angle Q M N$
$$
\begin{aligned}
\Leftrightarrow & \frac{P M}{Q M}=\frac{P N}{Q N} \\
\Leftrightarrow & \frac{x_{0}-x_{1}}{x_{0}-x_{2}}=-\frac{y_{1}}{y_{2}} \\
\Leftrightarrow & x_{0}=\frac{x_{1} y_{2}+x_{2} y_{1}}{y_{1}+y_{2}} \\
\quad & \frac{\left(k y_{1}+m\right) y_{2}+\left(k y_{2}+m\right) y_{1}}{y_{1}+y_{2}} \\
& =2 k \frac{y_{1} y_{2}}{y_{1}+y_{2}}+m \\
& =2 k \frac{A m^{2}+D m+F}{-(2 A m+D) k}+m \\
& =-\frac{D m+2 F}{2 A m+D},
\end{aligned}
$$

i.e., point $M\left(-\frac{D m+2 F}{2 A m+D}, 0\right)$.