8. Let $F_{1}, F_{2}$ 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 $\qquad$
4
8. 【Analysis and Solution】Let the lengths of the major axis, minor axis, and focal distance of the ellipse be $2a, 2b, 2c$, respectively. Then, from its equation, we know $a=3, b=2, c=\sqrt{5}$. Therefore, $\left|P F_{1}\right|+\left|P F_{2}\right|=2a=6$. Given that $\left|P F_{1}\right|:\left|P F_{2}\right|=2:1$, we can deduce that $\left|P F_{1}\right|=4$, $\left|P F_{2}\right|=2$. In $\triangle P F_{1} F_{2}$, the lengths of the three sides are $2, 4, 2\sqrt{5}$. Since $2^2 + 4^2 = (2\sqrt{5})^2$, it is evident that $\triangle P F_{1} F_{2}$ is a right triangle with the lengths of the two legs being 2 and 4. Hence, the area of $\triangle P F_{1} F_{2}$ is $\frac{1}{2}\left|P F_{1}\right| \cdot \left|P F_{2}\right| = \frac{1}{2} \times 2 \times 4 = 4$.