Example 4 (1999 National High School League Question) Given that point $P$ is 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 point $P$ is $\qquad$ .
Solve: Fill in $-\frac{64}{5}$. Reason: $a=4, b=3, c=\sqrt{a^{2}+b^{2}}=5$, the equation of the right directrix is $x=\frac{16}{5}$, foci $F_{1}(-5,0), F_{2}(5,0)$, eccentricity $e=\frac{c}{a}=\frac{5}{4}$. Let $P\left(x_{0}, y_{0}\right)$ be the distance to the right directrix $d$, if $P$ is on the right branch of the hyperbola, then $\left|P F_{1}\right|=\left|P F_{2}\right|+2 a=e d+2 a,\left|P F_{1}\right|+\left|P F_{2}\right|=(e d+2 a)+$ $e d=2 e d+2 a>2 d$, which contradicts that $d$ is the arithmetic mean of $\left|P F_{1}\right|,\left|P F_{2}\right|$, so $P$ is on the left branch of the hyperbola, then $\left|P F_{2}\right|-\left|P F_{1}\right|=2 a$. Also, from the given conditions, $\left|P F_{2}\right|+\left|P F_{1}\right|=2 d$, adding the two equations and dividing by 2 gives $\left|P F_{2}\right|=a+d$, also $\left|P F_{2}\right|=e d$, thus $e d=a+d, d=\frac{a}{e-1}=\frac{4}{\frac{5}{4}-1}=16$. Therefore, the x-coordinate of $P$ is $x_{0}=\frac{16}{5}-16=-\frac{64}{5}$.