$F_{1}, F_{2}$ 是粗圆 ${ }^{\frac{x^{2}}{9}}+\frac{y^{2}}{4}=1$ 的两个焦点, $\mathrm{p}$ 是椭圆上的点, 且 $\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1$, 则 $\triangle P F_{1} F_{2}$ 的面积为
答案: A

【解析】【分析】设 $\left|\mathrm{PF}_{2}\right|=\mathrm{x}$, 则 $\left|\mathrm{PF}_{1}\right|=2 \mathrm{x}$, 依题意, $\left|\mathrm{PF}_{1}\right|+\left|\mathrm{PF}_{2}\right|=\mathrm{x}+2 \mathrm{x}=3 \mathrm{x}=2 \mathrm{a}=6$, $\therefore \mathrm{x}=2,2 \mathrm{x}=4$,

即 $\left|\mathrm{PF}_{2}\right|=2,\left|\mathrm{PF}_{1}\right|=4$, 又 $\left|\mathrm{F}_{1} \mathrm{~F}_{2}\right|=2 \sqrt{9-4}=2 \sqrt{5}$,

$\therefore\left|P F_{1}\right|^{2}+\left|\mathrm{PF}_{2}\right|^{2}=\left|\mathrm{F}_{1} \mathrm{~F}_{2}\right|^{2}$,

$\therefore \triangle \mathrm{PF}_{1} \mathrm{~F}_{2}$ 为直角三角形,

$\therefore \triangle \mathrm{PF}_{1} \mathrm{~F}_{2}$ 的面积为 $\mathrm{S}=\frac{1}{2}\left|\mathrm{PF}_{1}\right|\left|\mathrm{PF}_{2}\right|=\frac{1}{2} \times 2 \times 4=4$.

故选 A.