双曲线左右焦点 F1MF2NMN123

8. 已知双曲线 $C:\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\ (a>0,b>0)$ 的左右焦点分别为 $F_{1}$、$F_{2}$，过 $F_{1}$ 的直线与曲线 $C$ 的左右两支分别交于点 $M$、$N$，且 $\left|F_{1}M\right|:\left|F_{2}N\right|:\left|MN\right|=1:2:3$，则曲线 $C$ 的离心率为（　　）

A. $2$　　B. $\dfrac{\sqrt{33}}{3}$　　C. $\dfrac{\sqrt{22}}{3}$　　D. $\dfrac{\sqrt{11}}{3}$

解： 依题设 $\left|F_{1}M\right|=m$，$\left|F_{2}N\right|=2m$，$\left|MN\right|=3m$

$\because\left|F_{1}N\right|-\left|F_{2}N\right|=2a$

$\therefore 4m-2m=2a$

$\therefore m=a$

$\therefore\left|F_{1}M\right|=a$，$\left|MN\right|=3a$，$\left|F_{2}N\right|=2a$

$\because\left|MF_{2}\right|-\left|MF_{1}\right|=2a$

$\therefore\left|MF_{2}\right|=3a$

$\therefore\cos\angle F_{1}MF_{2}=\dfrac{9a^{2}+a^{2}-4c^{2}}{2\cdot3a\cdot a}$，$\cos\angle NMF_{2}=\dfrac{9a^{2}+9a^{2}-4a^{2}}{2\cdot3a\cdot3a}$

$\because\cos\angle F_{1}MF_{2}=-\cos\angle NMF_{2}$

$\therefore 12c^{2}=44a^{2}$　$\therefore e=\dfrac{\sqrt{33}}{3}$

故选 B。