13 (12 points) Given the parabola $C: y^{2}=4 x$, construct a right-angled triangle $M A B$ inscribed in the parabola with $M(1,2)$ as the right-angle vertex.
(1) Prove that the line $A B$ passes through a fixed point;
(2) Draw a perpendicular from point $M$ to $A B$, intersecting $A B$ at point $N$. Find the equation of the locus of point $N$.
13 (1) Let the equation of line $AB$ be $x=my+n, A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}x=m y+n, \\ y^{2}=4 x,\end{array}\right.$ we get $y^{2}-4 m y-4 n=0$, so $y_{1}+y_{2}=4 m$, $y_{1} y_{2}=-4 n$. Since $\angle A M B=90^{\circ}$, we have $\overrightarrow{M A} \cdot \overrightarrow{M B}=0$, which means
$$
\left(x_{1}-1, y_{1}-2\right) \cdot\left(x_{2}-1, y_{2}-2\right)=0.
$$

Thus,
$$
\begin{array}{c}
\left(x_{1}-1\right)\left(x_{2}-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)=0. \\
\left(m y_{1}+n-1\right)\left(m y_{2}+n-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)=0, \\
\left(m^{2}+1\right) y_{1} y_{2}+(m n-m-2)\left(y_{1}+y_{2}\right)+(n-1)^{2}+4=0, \\
\left(m^{2}+1\right) \cdot(-4 n)+(m n-m-2) 4 m+n^{2}-2 n+5=0.
\end{array}
$$

Simplifying, we get $(n-3)^{2}=4(m+1)^{2}$, so $n-3=2(m+1)$ or $n-3=$ $-2(m+1)$.

When $n-3=2(m+1)$, i.e., $n=2 m+5$, the equation of line $AB$ is $x=$ $m(y+2)+5$ passing through the fixed point $P(5,-2)$;

When $n-3=-2(m+1)$, i.e., $n=-2 m+1$, the equation of line $AB$ is $x=m(y-2)+1$ passing through point $M(1,2)$, which is not valid.
Therefore, line $AB$ passes through the fixed point $P(5,-2)$.
(2) From (1), we know that the trajectory of point $N$ is a circle with $PM$ as its diameter (excluding the point (1, $\pm 2))$, whose equation is $(x-3)^{2}+y^{2}=8(x \neq 1)$.