4. 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. 4 Let the lengths of the major axis, minor axis, and focal distance of the ellipse be $2a, 2b, 2c$, respectively, then $a=3, b=2, c=\sqrt{5},\left|P F_{1}\right|+\left|P F_{2}\right|=$ $2a=6$, and $\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1$, so $\left|P F_{1}\right|=4,\left|P F_{2}\right|=2$. In $\triangle P F_{1} F_{2}$, the lengths of the three sides are 2, 4, and $2 \sqrt{5}$, and $2^{2}+4^{2}=(2 \sqrt{5})^{2}$, so $\triangle P F_{1} F_{2}$ is a right triangle. Therefore, $S_{\triangle P F_{1} F_{2}}=\frac{1}{2} \times 2 \times 4=4$.