1. Let $P$ be a moving point on the parabola $y^{2}=2 x$. Draw the tangent line of the parabola through $P$ and let it intersect the circle $\mathrm{e} O: x^{2}+y^{2}=1$ at points $M, N$. The tangents to $\mathrm{e} O$ at points $M, N$ intersect at point $Q$. Then the equation of the locus of point $Q$ is $\qquad$
1. $y^{2}=-2 x(x \leq 1-\sqrt{2})$

Analysis: Let $P\left(2 t^{2}, t\right)$, the equation of the tangent line to the parabola at point $P$ is: $2 y t=x+2 t^{2}$; if this tangent line intersects the unit circle, then $\frac{\left|2 t^{2}\right|}{\sqrt{t^{2}+1}} \leq 1$, which simplifies to $\left(2 t^{2}\right)^{2} \leq t^{2}+1$, solving this yields $0 \leq t^{2} \leq \frac{1+\sqrt{2}}{2}$. The intersection point $Q$ of the two tangents to the circle must be outside the circle. If we assume $Q\left(x_{0}, y_{0}\right)$, then the equation of the line $M N$ is: $x_{0} x+y_{0} y=1$, hence $x_{0}=-\frac{1}{2 t^{2}}, n=\frac{1}{t}$, thus the trajectory equation of point $Q$ is: $y^{2}=-2 x$ and $x=-\frac{1}{2 t^{2}} \leq 1-\sqrt{2}$.