已知 $F_{1}, F_{2}$ 是双曲线 $E: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ 的左、右焦点, 点 $M$ 在 $E$ 上, $M F_{1}$ 与 $x$ 轴垂直, $\sin \angle M F_{2} F_{1}=\frac{1}{3}$,则 $E$ 的离心率为 ( )
设 $\left|M F_{1}
ight|=m$, 则 $\left|M F_{2}
ight|=2 a+m, ecause M F_{1}$ 与 $x$ 轴垂直, $	herefore(2 a+m)^{2}=m^{2}+4 c^{2}$, $	herefore m=rac{b^{2}}{a}: \sin \angle M F_{2} F_{1}=rac{1}{3}, \quad 	herefore rac{b^{2}}{a}=a=m, \quad 	herefore a=b, \quad 	herefore c=\sqrt{2} a, \quad 	herefore e=rac{c}{a}=\sqrt{2}$.