设 $P$ 为粗圆 $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ 上的一点, $F_{1} 、 F_{2}$ 是该粗圆的两个焦点, 若 $\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1$ 则 $\triangle P F_{1} F_{2}$ 的面积为 ( $)$
$ecause\left|P F_{1}
ight|:\left|P F_{2}
ight|=2: 1, 	herefore$ 可设 $\left|P F_{1}
ight|=2 k,\left|P F_{2}
ight|=k$, 由题意可知 $2 k+k=6, 	herefore k=2, 	herefore\left|P F_{1}
ight|=4$, $\left|P F_{2}
ight|=2, ecause\left|F_{1} F_{2}
ight|=2 \sqrt{5}, 	herefore 	riangle P F_{1} F_{2}$ 是直角三角形, 其面积 $=rac{1}{2} 	imes\left|P F_{1}
ight| 	imes\left|P F_{2}
ight|=rac{1}{2} 	imes 4 	imes 2=4$. 故选 $C$.