一道秋双曲线离心率的数学试题

已知圆 $C_{1}:x^{2}+\left(y-\dfrac{2\sqrt{3}}{3}\right)^{2}=\dfrac{16}{3}$ 过双曲线 $C_{2}:\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\ (a>0,b>0)$ 的左右焦点 $F_{1}$，$F_{2}$，曲线 $C_{1}$ 与曲线 $C_{2}$ 在第一象限的交点为 $M$，$\left|MF_{1}\right|\cdot\left|MF_{2}\right|=12$，则双曲线的离心率为__。

解： $\because$ 在直角 $\triangle F_{1}C_{1}O$ 中，$\left|C_{1}F_{1}\right|=\dfrac{4\sqrt{3}}{3}$，$\left|C_{1}O\right|=\dfrac{2\sqrt{3}}{3}$

$\therefore\left|F_{1}O\right|=c=2$，$\cos F_{1}C_{1}O=\dfrac{1}{2}$

设 $\left|MF_{1}\right|=m$，$\left|MF_{2}\right|=n$

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

$\therefore m-n=2a$，$mn=12$

$\therefore m^{2}+n^{2}=4a^{2}+24$

$\therefore\cos\angle F_{1}MF_{2}=\dfrac{m^{2}+n^{2}-4c^{2}}{2mn}=\dfrac{4a^{2}+24-16}{24}=\dfrac{a^{2}+2}{6}$

$\because\angle F_{1}C_{1}F_{2}=2F_{1}C_{1}O=2\angle F_{1}MF_{2}$

$\therefore\cos F_{1}C_{1}O=\cos\angle F_{1}MF_{2}$

$\therefore\dfrac{a^{2}+2}{6}=\dfrac{1}{2}$

$\therefore a=1$

$\therefore e=\dfrac{c}{a}=4$