10. (20 points) Given the ellipse $\Gamma: \frac{x^{2}}{2}+y^{2}=1$ with its left focus at $F$, a line is drawn from a point $P$ outside the ellipse to be tangent to $\Gamma$ at points $M$ and $N$. If $\angle M F N=60^{\circ}$, find the equation of the locus of point $P$.


10. From $\angle M F N=60^{\circ}$, we know
$$
\frac{1}{2}=\cos \angle M F N=\frac{\overrightarrow{F M} \cdot \overrightarrow{F N}}{|\overrightarrow{F M}||\overrightarrow{F N}|} \text {. }
$$

Let point $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Since $F(-1,0)$, the eccentricity $e=\sqrt{\frac{1}{2}}$, and the left directrix $x=-2$, then
$$
\begin{array}{l}
\frac{1}{2}=\frac{\overrightarrow{F M} \cdot \overrightarrow{F N}}{|\overrightarrow{F M}||\overrightarrow{F N}|} \\
=\frac{\left(x_{1}+1\right)\left(x_{2}+1\right)+y_{1} y_{2}}{\frac{1}{2}\left(x_{1}+2\right)\left(x_{2}+2\right)} \\
\Rightarrow 3 x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4 y_{1} y_{2}=0 .
\end{array}
$$

Let point $P\left(x_{0}, y_{0}\right)$. Then the equation of line $M N$ is
$$
\frac{x_{0} x}{2}+y_{0} y=1 \text {, }
$$

combined with the equation of the ellipse $\Gamma$ we get
$$
\left(x_{0}^{2}+2 y_{0}^{2}\right) x^{2}-4 x_{0} x+4-4 y_{0}^{2}=0 \text {. }
$$

By Vieta's formulas, we have
$$
x_{1} x_{2}=\frac{4-4 y_{0}^{2}}{x_{0}^{2}+2 y_{0}^{2}}, x_{1}+x_{2}=\frac{4 x_{0}}{x_{0}^{2}+2 y_{0}^{2}} \text {. }
$$

Substituting into equation (1), we solve to get
$$
y_{1} y_{2}=\frac{2-x_{0}^{2}}{x_{0}^{2}+2 y_{0}^{2}} \text {. }
$$

Thus, $3\left(4-4 y_{0}^{2}\right)+2 \times 4 x_{0}+4\left(2-x_{0}^{2}\right)=0$.
Therefore, the trajectory of point $P$ is
$$
\frac{(x-1)^{2}}{6}+\frac{y^{2}}{2}=1 \text {. }
$$