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 I: 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 $\mathrm{J}$
From the definition of an ellipse, we know that $\left|P F_{1}\right|+\left|P F_{2}\right|=2 a=6$, and $\left|P F_{1}\right|$ : $\left|P F_{2}\right|=2: 1$. Therefore, $\left|P F_{1}\right|=4 \cdot\left|P F_{2}\right|=2$, and $\left|F_{1} F_{2}\right|=2 c=$ $2 \sqrt{5}$. Combining $4^{2}+2^{2}=(2 \sqrt{5})^{2}$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, hence
$$
S_{\triangle P F_{1} F_{0}}=\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right|=4 .
$$