8. Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and $P$ be a point on the ellipse such that $\left|P F_{1}\right|:\left|P F_{2}\right|$ $=2: 1$. Then the area of $\triangle P F_{1} F_{2}$ is $\qquad$.
Given $\left\{\begin{array}{l}\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1, \\ \left|P F_{1}\right|+\left|P F_{2}\right|=6\end{array} \Rightarrow\left\{\begin{array}{l}\left|P F_{1}\right|=4, \\ \left|P F_{2}\right|=2 .\end{array}\right.\right.$ Also, $\left|F_{1} F_{2}\right|=2 \sqrt{5}$,
so $\angle F_{1} P F_{2}=90^{\circ} \Rightarrow S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right|=4$.