13. 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 $N$.
(1) Let $AB: x=ny+m, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, combining $\left\{\begin{array}{l}x=ny+m, \\ y^{2}=4x\end{array}\right.$ $\Rightarrow y^{2}-4ny-4m=0$. Thus, $\left(x_{1}-1\right)\left(x_{2}-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)$ $=x_{1}x_{2}-\left(x_{1}+x_{2}\right)+y_{1}y_{2}-2\left(y_{1}+y_{2}\right)+5=\frac{y_{1}^{2}y_{2}^{2}}{16}-\frac{1}{4}\left(y_{1}+y_{2}\right)^{2}+\frac{3y_{1}y_{2}}{2}-2\left(y_{1}+y_{2}\right)+5$ $=m^{2}-4n^{2}-6m-8n+5=(m+2n-1)(m-2n-5)=0$.
When $m=-2n+1$, the line $AB$ always passes through the fixed point $M(1,2)$, which does not meet the problem's requirement, so it is discarded; when $m=2n+5$, the line $AB$ always passes through the fixed point $P(5,-2)$.
(2) Note that $M(1,2), P(5,-2)$ are both fixed points, and $\angle MNP=90^{\circ}$, then point $N$ is always on the circle with $MP$ as its diameter, but excluding the points $(1,2),(1,-2)$. Therefore, the trajectory equation of point $N$ is $(x-3)^{2}+y^{2}=8(x \neq 1)$.