10. From any point $P$ on the parabola $y^{2}=2 x$, draw a perpendicular to its directrix $l$, with the foot of the perpendicular being $Q$. The line connecting the vertex $O$ and $P$ and the line connecting the focus $F$ and $Q$ intersect at point $R$. Then the equation of the locus of point $R$ is $\qquad$.
10. $y^{2}=-2 x^{2}+x$. Hint: Let $P\left(x_{1}, y_{1}\right), R(x, y)$, then $Q\left(-\frac{1}{2}, y_{1}\right), F\left(\frac{1}{2}, 0\right)$, then the equation of $O P$ is: $y=\frac{y_{1}}{x_{1}} x, F Q$'s equation is: $y=-y_{1}\left(x-\frac{1}{2}\right)$, eliminate $x_{1}, y_{1}$ (substitution method) and you are done.