10. (20 points) From a point outside the parabola $y^{2}=2 p x(p>0)$, draw two tangent lines $P A, P B$ to the parabola, with points of tangency at $A, B$ respectively. Let the circumcenter of $\triangle P A B$ be $Q$, and $Q$ does not coincide with the focus $F$ of the parabola. Try to find $\angle P F Q$.
10. Let $A\left(2 p t_{1}^{2}, 2 p t_{1}\right), B\left(2 p t_{2}^{2}, 2 p t_{2}\right)$. It is easy to know that the line $P A: 2 t_{1} y=x+2 p t_{1}^{2}, P B: 2 t_{2} y=x+2 p t_{2}^{2}$.

Solving these equations simultaneously, we get $P\left(2 p t_{1} t_{2}, p\left(t_{1}+t_{2}\right)\right)$. Further, we can write the equations of the perpendicular bisectors $l_{1}, l_{2}$ of segments $P A, P B$ as
$$
\begin{array}{l}
l_{1}: y+2 t_{1} x=2 p t_{1}^{2}\left(t_{1}+t_{2}\right)+\frac{p\left(3 t_{1}+t_{2}\right)}{2} \\
l_{2}: y+2 t_{2} x=2 p t_{2}^{2}\left(t_{1}+t_{2}\right)+\frac{p\left(3 t_{2}+t_{1}\right)}{2} .
\end{array}
$$

Solving $l_{1}, l_{2}$ simultaneously, we get the coordinates of point $Q$ as
$$
\left(p\left(t_{1}+t_{2}\right)^{2}+\frac{p}{2},-2 p t_{1} t_{2}\left(t_{1}+t_{2}\right)+\frac{p\left(t_{1}+t_{2}\right)}{2}\right) .
$$

Combining with $F\left(\frac{p}{2}, 0\right)$, we have
$$
\begin{array}{l}
\overrightarrow{F P}=\left(p\left(2 t_{1} t_{2}-\frac{1}{2}\right), p\left(t_{1}+t_{2}\right)\right) \\
\overrightarrow{F Q}=\left(p\left(t_{1}+t_{2}\right)^{2},-2 p t_{1} t_{2}\left(t_{1}+t_{2}\right)+\frac{p\left(t_{1}+t_{2}\right)}{2}\right)
\end{array}
$$

It is easy to see that $\overrightarrow{F P} \cdot \overrightarrow{F Q}=0$, hence $\angle P F Q=90^{\circ}$.