4. Let the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have a focus at $F$. Draw a line $l$ through $F$ perpendicular to the $x$-axis, intersecting the two asymptotes at points $A$ and $B$. Let $P$ be one of the points where $l$ intersects the hyperbola. Let $O$ be the origin. If there exist real numbers $m$ and $n$ such that $\overrightarrow{O P}=m \overrightarrow{O A}+n \overrightarrow{O B}$, and $m n=\frac{2}{9}$, then the eccentricity of the hyperbola is ( ).
(A) $\frac{3 \sqrt{2}}{4}$
(B) $\frac{9}{8}$
(C) $\frac{3 \sqrt{5}}{5}$
(D) $\frac{3 \sqrt{2}}{2}$
4. A.

Substitute the coordinates of points $A\left(c, \frac{b c}{a}\right), B\left(c,-\frac{b c}{a}\right)$ into
$$
\overrightarrow{O P}=m \overrightarrow{O A}+n \overrightarrow{O B} \text {, }
$$

to get point $P\left((m+n) c,(m-n) \frac{b c}{a}\right)$.
Substitute the coordinates of point $P$ into the hyperbola equation to get
$$
\begin{array}{l}
\frac{(m+n)^{2} c^{2}}{a^{2}}-\frac{(m-n)^{2} b^{2} c^{2}}{a^{2} b^{2}}=1 \\
\Rightarrow 4 m n c^{2}=a^{2} \\
\Rightarrow e=\frac{c}{a}=\frac{1}{\sqrt{4 m n}}=\sqrt{\frac{9}{8}}=\frac{3 \sqrt{2}}{4} .
\end{array}
$$