13. Draw two perpendicular chords $O A, O B$ through the vertex of the parabola $y^{2}=4 p x(p>0)$, find the locus of the projection $M$ of the vertex $O$ of the parabola on the line $A B$.
13. Let the equation of $O A$ be $y=k x$, then the equation of $O B$ is $y=-\frac{x}{k}$, from which we can find $A\left(\frac{4 p}{k^{2}}, \frac{4 p}{k}\right), B\left(4 p k^{2},-4 k p\right)$. Thus, the slope of $A B$ is $k_{A B}=\frac{k}{1-k^{2}}$, hence $k_{O M}=\frac{k^{2}-1}{k}$, so the equation of $O M$ is $y=\frac{k^{2}-1}{k} x$. (1) $\quad A B$'s equation is $y+4 p k=\frac{-k}{k^{2}-1}\left(x-4 p k^{2}\right)$.
(2) (1) $\times$ (2) gives $y^{2}+4 p k y=-x\left(x-4 p k^{2}\right)$. That is, $x^{2}+y^{2}=-4 p k y+4 p k^{2} x$ $=4 p\left(k^{2} x-k y\right)$.
(3) From (1), we get $k^{2} x-k y=x$ substituting into
(3) and simplifying yields $(x-2 p)^{2}+y^{2}=4 p^{2}$, so the trajectory equation of point $M$ is $(x-2 p)^{2}+y^{2}=4 p^{2}$, which is a circle with center $(2 p, 0)$ and radius $2 p$.