4. Given that point $P$ lies on the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, and the distance from $P$ to the right directrix of this hyperbola is exactly the arithmetic mean of the distances from $P$ to the two foci of this hyperbola, then the x-coordinate of $P$ is $\qquad$
$$
-\frac{64}{5}
$$
4. 【Analysis and Solution】Let the lengths of the semi-real axis, semi-imaginary axis, and semi-focal distance be $a$, $b$, and $c$ respectively, the eccentricity be $e$, and the distance from point $P$ to the right directrix $l$ be $d$. Then $a=4$, $b=3$, $c=5$, $e=\frac{c}{a}=\frac{5}{4}$, and the right directrix $l$ is $x=\frac{a^{2}}{c}=\frac{16}{5}$.
If $P$ is on the right branch of the hyperbola, then $\left|P F_{1}\right|=\left|P F_{2}\right|+2 a=e d+2 a$,
thus, $\left|P F_{1}\right|+\left|P F_{2}\right|=(e d+2 a)+e d=2 e d+2 a>2 d$, which is impossible.
Therefore, $P$ is on the left branch of the hyperbola, and $\left|P F_{2}\right|-\left|P F_{1}\right|=2 a$, $\left|P F_{2}\right|+\left|P F_{1}\right|=2 d$,
adding these two equations, we get $2\left|P F_{2}\right|=2 a+2 d$. Also, $\left|P F_{2}\right|=e d$,
$\therefore d=\frac{a}{e-1}=\frac{4}{\frac{5}{4}-1}=16$, hence, the x-coordinate of $P$ is $\frac{16}{5}-16=-\frac{64}{5}$.