Question 175, Given point $E(m, n)$ is a fixed point inside the parabola $y^{2}=2 p x(p>0)$, two lines with slopes $k_{1}$ and $k_{2}$ are drawn through $E$, intersecting the parabola at points $A, B$ and $C, D$ respectively, and $M, N$ are the midpoints of segments $A B$ and $C D$ respectively.
(1) When $\mathrm{n}=0$, and $\mathrm{k}_{1} \cdot \mathrm{k}_{2}=-1$, find the minimum value of the area of $\triangle E M N$.
(2) If $k_{1}+\mathrm{k}_{2}=\lambda(\lambda \neq 0$, and $\lambda$ is a constant), prove that the line $\mathbb{N N}$ passes through a fixed point.
Question 175, Solution: (1) The equation of line $AB$ is $y=k_{1}(x-m)$. By solving the system of equations $\left\{\begin{array}{c}y^{2}=2 p x \\ y=k_{1}(x-m)\end{array}\right.$, we get $k_{1}^{2}(x-m)^{2}=2 p x$. Rearranging, we have $k_{1}^{2} x^{2}-2\left(k_{1}^{2} m+p\right) x+k_{1}^{2} m^{2}=0$. Thus, the x-coordinates of points $A$ and $B$, $x_{A}$ and $x_{B}$, are the roots of the equation $k_{1}^{2} x^{2}-2\left(k_{1}^{2} m+p\right) x+k_{1}^{2} m^{2}=0$. According to Vieta's formulas, $x_{M}=\frac{x_{A}+x_{B}}{2}=\frac{k_{1}^{2} m+p}{k_{1}^{2}}$. Therefore, $|E M|=\sqrt{1+k_{1}^{2}} \cdot\left|\frac{k_{1}^{2} m+p}{k_{1}^{2}}-m\right|=\sqrt{1+k_{1}^{2}} \cdot\left|\frac{p}{k_{1}^{2}}\right|$. Similarly, $|E N|=\sqrt{1+k_{2}^{2}} \cdot\left|\frac{p}{k_{2}^{2}}\right|$. Thus, using the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
S_{\triangle E M N}=\frac{1}{2} \cdot|E M| \cdot|E N|=\frac{1}{2} \sqrt{\left(1+k_{1}^{2}\right)\left(1+k_{2}^{2}\right)} \cdot \frac{p^{2}}{\left(k_{1} \cdot k_{2}\right)^{2}} \\
\geq \frac{1}{2}\left(1+\left|k_{1} \cdot k_{2}\right|\right) \cdot \frac{p^{2}}{\left(k_{1} \cdot k_{2}\right)^{2}}=p^{2}
\end{array}
$$

Therefore, the minimum area of $\triangle E M N$ is $p^{2}$.
(2) The equation of line $AB$ is $y=k_{1}(x-m)+n$. By solving the system of equations $\left\{\begin{array}{c}y^{2}=2 p x \\ y=k_{1}(x-m)+n\end{array}\right.$, we get $\left[k_{1}(x-m)+n\right]^{2}=2 p x$. Rearranging, we have $k_{1}^{2} x^{2}-2\left(k_{1}^{2} m+p-k_{1} n\right) x+\left(k_{1} m-n\right)^{2}=0$. According to Vieta's formulas, $x_{M}=\frac{x_{A}+x_{B}}{2}=\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}$, and thus $y_{M}=k_{1}\left(\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}-m\right)+n=\frac{p}{k_{1}}$. Therefore, point $M\left(\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}, \frac{p}{k_{1}}\right)$. Similarly, point $N\left(\frac{k_{2}^{2} m+p-k_{2} n}{k_{2}^{2}}, \frac{p}{k_{2}}\right)$. Thus, the equation of line $MN$ is: $y-\frac{p}{k_{1}}=$ $\frac{\frac{p}{k_{2} k_{1}}}{\left(\frac{k_{2} m+p-k_{2} n}{k_{2}}\right)-\left(\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}\right)} \cdot\left(x-\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}\right)$, rearranging, we get:
$$
y=\frac{\frac{p}{k_{2}} k_{1}}{\left(\frac{k_{2}^{2} m+p-k_{2} n}{k_{2}^{2}}\right)-\left(\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}\right)} \cdot\left(x-\frac{k_{1}^{2} m+p-k_{1} n}{k_{1}^{2}}\right)+\frac{p}{k_{1}}
$$
$$
\begin{array}{l}
=\frac{k_{1} k_{2} p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}\left(x-m-\frac{p-k_{1} n}{k_{1}^{2}}\right)+\frac{p}{k_{1}} \\
=\frac{k_{1} k_{2} p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}(x-m)-\frac{k_{2} p-k_{1} k_{1} n}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n} \cdot \frac{p}{k_{1}}+\frac{p}{k_{1}} \\
=\frac{k_{1} k_{2} p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}(x-m)+\frac{p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n} \\
=\frac{p\left(k_{1}\right.}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}\left(k_{1} k_{2} x-k_{1} k_{2} m+p\right) \\
=\frac{p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}\left[k_{1} k_{2} x-k_{1} k_{2} m+\frac{k_{1} k_{2} n}{k_{1}+k_{2}}+\left(p-\frac{k_{1} k_{2} n}{k_{1}+k_{2}}\right)\right] \\
=\frac{p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}\left[k_{1} k_{2} x-k_{1} k_{2} m+\frac{k_{1} k_{2} n}{k_{1}+k_{2}}+\left(\frac{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}{k_{1}+k_{2}}\right)\right] \\
=\frac{k_{1} k_{2} p}{p\left(k_{1}+k_{2}\right)-k_{1} k_{2} n}\left(x-m+\frac{n}{k_{1}+k_{2}}\right)+\frac{p}{k_{1}+k_{2}} \\
=\frac{k_{1} k_{2} p}{p-k_{1} k_{2} n}\left[x-\left(m-\frac{n}{\lambda}\right)\right]+\frac{p}{\lambda}
\end{array}
$$

Thus, line $MN$ always passes through the point $\left(m-\frac{n}{\lambda}, \frac{p}{\lambda}\right)$.