9. (16 points) Let the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ have its right focus at $F$, and let $P$ be a point outside the hyperbola. Draw two tangents from point $P$ to the hyperbola, touching it at points $A$ and $B$. If $A B \perp P F$, find the locus of point $P$.
9. Let point $P\left(x_{0}, y_{0}\right)$, tangent points $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and the right focus $F\left(\sqrt{a^{2}+b^{2}}, 0\right)$.
By the chord of contact formula, we have $l_{A B}: \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$.
When $y_{0}=0$, by symmetry, points $A$ and $B$ are symmetric with respect to the $x$-axis, and $A B \perp P F$, satisfying the condition. In this case, the locus of point $P$ is the line segment $y=0(-a<x<a)$.
When $y_{0} \neq 0$,
$$
k_{A B}=\frac{b^{2} x_{0}}{a^{2} y_{0}}, k_{P F}=\frac{y_{0}}{x_{0}-\sqrt{a^{2}+b^{2}}} .
$$

By $A B \perp P F$
$\Rightarrow k_{A B} k_{P F}=\frac{b^{2} x_{0}}{a^{2} y_{0}} \cdot \frac{y_{0}}{x_{0}-\sqrt{a^{2}+b^{2}}}=-1$
$$
\Rightarrow x_{0}=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}} \text {. }
$$

In summary, the locus of point $P$ is the line $x=\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$ and the line segment $y=0(-a<x<a)$.