9. (16 points) Two tangents are drawn from point $P$ to the ellipse $4 x^{2}+5 y^{2}=20$ and they are perpendicular to each other. Find the equation of the locus of point $P$.


9. As shown in Figure 3, let the two tangents touch the ellipse at points $S$ and $T$, and the symmetric points of $F_{2}$ about $PS$ and $PT$ be $P_{1}$ and $P_{2}$, respectively.

By the well-known optical property of ellipses, we have $F_{1}, S, P_{1}$ and $F_{1}, T, P_{2}$ are collinear, respectively.
Thus, $F_{1} P_{1} = F_{1} P_{2} = 2a$.
Also, $\angle SPT = 90^{\circ}$, so $P_{1}P_{2}$ has $P$ as its midpoint.
$$
\begin{array}{l}
\text{Therefore, } P F_{1}^{2} + P F_{2}^{2} = P F_{1}^{2} + P P_{1}^{2} = F_{1} P_{1}^{2} = 4a^{2} \\
\Rightarrow \left(x_{0} + c\right)^{2} + y_{0}^{2} + \left(x_{0} - c\right)^{2} + y_{0}^{2} = 4a^{2} \\
\Rightarrow x_{0}^{2} + y_{0}^{2} = a^{2} + b^{2}.
\end{array}
$$

Therefore, the equation of the locus of point $P$ is $x^{2} + y^{2} = 9$.