11. (20 points) Given $E(m, n)$ is a fixed point inside the parabola
$$
y^{2}=2 p x(p>0)
$$

two lines passing through $E$ with slopes $k_{1}$ and $k_{2}$ intersect the parabola at points $A, B, C, D$, and $M, N$ are the midpoints of segments $AB$ and $CD$, respectively.
(1) When $n=0$ and $k_{1} k_{2}=-1$, find the minimum value of the area of $\triangle EMN$;
(2) If $k_{1}+k_{2}=\lambda($ constant $\lambda \neq 0)$, prove that the line $MN$ passes through a fixed point.
11. Let the line $l_{A B}: x=t_{1}(y-n)+m\left(t_{1}=\frac{1}{k_{1}}\right)$.

Substitute into $y^{2}=2 p x$, we get
$$
y^{2}-2 p t_{1} y+2 p t_{1} n-2 p m=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $y_{1}+y_{2}=2 p t_{1}$
$$
\begin{aligned}
\Rightarrow & x_{1}+x_{2}=t_{1}\left(y_{1}+y_{2}-2 n\right)+2 m \\
& =t_{1}\left(2 p t_{1}-2 n\right)+2 m .
\end{aligned}
$$

Thus, $M\left(p t_{1}^{2}-n t_{1}+m, p t_{1}\right)$.
Similarly,
Let the line $l_{C D}: x=t_{2}(y-n)+m\left(t_{2}=\frac{1}{k_{2}}\right)$, we get $N\left(p t_{2}^{2}-n t_{2}+m, p t_{2}\right)$.
(1) When $n=0$,
$$
\begin{array}{l}
E(m, 0), M\left(p t_{1}^{2}+m, p t_{1}\right), N\left(p t_{2}^{2}+m, p t_{2}\right), \\
|E M|=\left|p t_{1}\right| \sqrt{1+t_{1}^{2}}, \\
|E N|=\left|p t_{2}\right| \sqrt{1+t_{2}^{2}} .
\end{array}
$$

Also, $k_{1} k_{2}=-1$, then $t_{1} t_{2}=-1$.
Thus, $S_{\triangle E M N}=\frac{1}{2}|E M||E N|$
$$
\begin{array}{l}
=\frac{1}{2}\left|p^{2} t_{1} t_{2}\right| \sqrt{\left(1+t_{1}^{2}\right)\left(1+t_{2}^{2}\right)} \\
=\frac{p^{2}}{2} \sqrt{2+t_{1}^{2}+t_{2}^{2}} \geqslant \frac{p^{2}}{2} \sqrt{4}=p^{2} .
\end{array}
$$

The equality holds if and only if $\left|t_{1}\right|=\left|t_{2}\right|=1$. Therefore, the minimum area of $\triangle E M N$ is $p^{2}$.
(2) Note that,
$$
\begin{array}{l}
k_{M N}=\frac{p\left(t_{1}-t_{2}\right)}{p\left(t_{1}^{2}-t_{2}^{2}\right)-n\left(t_{1}-t_{2}\right)} \\
=\frac{1}{\left(t_{1}+t_{2}\right)-\frac{n}{p}} .
\end{array}
$$

Thus, the line $l_{M N}$ :
$$
\begin{array}{l}
y-p t_{1}=\frac{1}{\left(t_{1}+t_{2}\right)-\frac{n}{p}}\left(x-\left(p t_{1}^{2}-n t_{1}+m\right)\right) \\
\Rightarrow y\left(t_{1}+t_{2}-\frac{n}{p}\right)-p t_{1} t_{2}=x-m .
\end{array}
$$

Also, $k_{1}+k_{2}=\frac{1}{t_{1}}+\frac{1}{t_{2}}=\lambda$, i.e., $t_{1} t_{2}=\frac{t_{1}+t_{2}}{\lambda}$, substituting into
equation (1) we get
$$
\begin{array}{l}
y\left(t_{1}+t_{2}-\frac{n}{p}\right)-p \cdot \frac{t_{1}+t_{2}}{\lambda}=x-m \\
\Rightarrow\left(t_{1}+t_{2}\right)\left(y-\frac{p}{\lambda}\right)=x+\frac{n y}{p}-m .
\end{array}
$$

When $y-\frac{p}{\lambda}=0$, we have $x+\frac{n y}{p}-m=0$, i.e., $\left(m-\frac{n}{\lambda}, \frac{p}{\lambda}\right)$ is a solution to the equation.
Therefore, the line $M N$ always passes through the fixed point $\left(m-\frac{n}{\lambda}, \frac{p}{\lambda}\right)$.