11 Given point $E(m, n)$ 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 $A B$ and $C D$, respectively.
(1) When $n=0$ and $k_{1} \cdot k_{2}=-1$, find the minimum value of the area of $\triangle E M N$;
(2) If $k_{1}+k_{2}=\lambda(\lambda \neq 0, \lambda$ is a constant), prove that the line $M N$ passes through a fixed point.
The equation of the line on which $11 A B$ lies is $x=t_{1}(y-n)+m$, where $t_{1}=\frac{1}{k_{1}}$. Substituting into $y^{2}=2 p x$, we get
$$
y^{2}-2 p t_{1} y+2 p t_{1} n-2 p m=0.
$$

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then $y_{1}+y_{2}=2 p t_{1}$, and thus
$$
x_{1}+x_{2}=t_{1}\left(y_{1}+y_{2}-2 n\right)+2 m=t_{1}\left(2 p t_{1}-2 n\right)+2 m,
$$

so $M\left(p t_{1}^{2}-n t_{1}+m, p t_{1}\right)$. The equation of the line on which $C D$ lies is $x=t_{2}(y-n)+m$, where $t_{2}=\frac{1}{k_{2}}$. Similarly, we get $N\left(p t_{2}^{2}-n t_{2}+m, p t_{2}\right)$.
(1) When $n=0$, $E(m, 0)$, $M\left(p t_{1}^{2}+m, p t_{1}\right)$, $N\left(p t_{2}^{2}+m, p t_{2}\right)$, $|E M|=\left|p t_{1}\right| \sqrt{1+t_{1}^{2}}$, $|E N|=\left|p t_{2}\right| \sqrt{1+t_{2}^{2}}$.
Since $k_{1} \cdot k_{2}=-1$, we have $t_{1} \cdot t_{2}=-1$. Therefore, the area of $\triangle E M N$ is
$$
\begin{aligned}
S & =\frac{1}{2}|E M| \cdot|E N| \\
& =\frac{1}{2}\left|p^{2} t_{1} t_{2}\right| \sqrt{\left(1+t_{1}^{2}\right)\left(1+t_{2}^{2}\right)} \\
& =\frac{p^{2}}{2} \cdot \sqrt{2+t_{1}^{2}+t_{2}^{2}} \\
& \geqslant \frac{p^{2}}{2} \cdot \sqrt{4}=p^{2},
\end{aligned}
$$

with equality holding if and only if $\left|t_{1}\right|=\left|t_{2}\right|=1$.
Thus, the minimum area of $\triangle E M N$ is $p^{2}$.
(2) $k_{M N}=\frac{p\left(t_{1}-t_{2}\right)}{p\left(t_{1}^{2}-t_{2}^{2}\right)-n\left(t_{1}-t_{2}\right)}=\frac{1}{\left(t_{1}+t_{2}\right)-\frac{n}{p}}$.
The equation of the line $M N$ is
$$
y-p t_{1}=\frac{1}{\left(t_{1}+t_{2}\right)-\frac{n}{p}} \cdot\left[x-\left(p t_{1}^{2}-n t_{1}+m\right)\right],
$$

which simplifies to
$$
y\left(t_{1}+t_{2}-\frac{n}{p}\right)-p t_{1} t_{2}=x-m.
$$

Since $k_{1}+k_{2}=\frac{1}{t_{1}}+\frac{1}{t_{2}}=\lambda$, we have $t_{1} t_{2}=\frac{t_{1}+t_{2}}{\lambda}$. Substituting this into the equation, we get
$$
y\left(t_{1}+t_{2}-\frac{n}{p}\right)-p \cdot \frac{t_{1}+t_{2}}{\lambda}=x-m,
$$

which simplifies to
$$
\left(t_{1}+t_{2}\right)\left(y-\frac{p}{\lambda}\right)=x+\frac{n y}{p}-m.
$$

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