9. (14 points) Let $P$ move on the right branch of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ (excluding the two vertices on the major axis), and let points $E$ and $F$ be the left and right foci, respectively. $A$ is the excenter of $\triangle P E F$ with respect to $\angle P E F$ (the intersection of the internal angle bisector of $\angle P E F$ and the external angle bisectors of $\angle P E F$ and $\angle E P F$).
Find the equation of the trajectory of point $A$.
9. Solution: Let the coordinates of $A$ and $P$ be $A(x, y)$ and $P\left(x_{0}, y_{0}\right)$, respectively, then $x_{0}>4$.
Let $P A$ intersect the $x$-axis at point $B$.
By the property of the angle bisector, we have $\frac{|B A|}{|A P|}=\frac{|B E|}{|P E|}=\frac{|B F|}{|P F|}=$ $\frac{|B E|-|B F|}{|P E|-|P F|}=\frac{2 c}{2 a}=\frac{5}{4}$.
Thus, point $A$ divides the directed line segment $\overrightarrow{B P}$ in the ratio $\lambda=\frac{5}{4}$. By the section formula, we get
$$
\left\{\begin{array}{l}
x=\frac{x_{B}+\lambda x_{0}}{1+\lambda}=\frac{4 x_{B}+5 x_{0}}{9} \\
y=\frac{y_{B}+\lambda y_{0}}{1+\lambda}=\frac{5 y_{0}}{9} .
\end{array} \cdots \cdots(*) .\right.
$$

By the focal radius formula of the hyperbola, we get $|P F|=e x_{0}-a=$ $\frac{5}{4} x_{0}-4$.
And $|B F|=x_{B}-5$, then $\frac{x_{B}-5}{\frac{5}{4} x_{0}-4}=\frac{5}{4}$,
which gives $x_{B}=\frac{25}{15} x_{0}$.
Thus, from (*) we have $\left\{\begin{array}{l}x_{0}=\frac{4}{5} x \\ y_{0}=\frac{9}{5} y\end{array}\right.$.
Since $P\left(x_{0}, y_{0}\right)$ lies on the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, we have $\frac{x_{0}^{2}}{16}-\frac{y_{0}^{2}}{9}=1$,
hence the equation of the locus of point $A$ is $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1$.
Therefore, when $P$ is on the right branch of the hyperbola, the required equation of the locus of point $A$ is $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5)$.