离心率 双曲线e2+a2最小值

15. 已知双曲线 $C:\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\ (a>0,b>0)$ 的右焦点为 $F$，离心率为 $e$，过原点的直线与 $C$ 的左右两支分别交于 $M,N$ 两点，若 $\left|MF\right|\left|NF\right|=4$，$\angle MFN=60^{\circ}$，则 $e^{2}+\dfrac{a^{2}}{4}$ 的最小值为__。

解： 依题可知四边形 $MNFF_{1}$ 为平行四边形

由余弦定理可得

$\left|MN\right|^{2}=\left|MF\right|^{2}+\left|NF\right|^{2}-2\left|MF\right|\left|NF\right|\cos 60^{\circ}$

$\left|F_{1}F\right|^{2}=\left|MF_{1}\right|^{2}+\left|MF\right|^{2}-2\left|MF_{1}\right|\left|MF\right|\cos 120^{\circ}$

$\therefore\left|MN\right|^{2}=\left|MF\right|^{2}+\left|NF\right|^{2}-4$

$4c^{2}=\left|MF_{1}\right|^{2}+\left|MF\right|^{2}+4$

$\therefore\left|MN\right|^{2}=4c^{2}-8$

$\because\left|MF\right|-\left|NF\right|=2a$

$\therefore\left|MF\right|^{2}+\left|NF\right|^{2}-2\left|MF\right|\left|NF\right|=4a^{2}$

$\therefore\left|MN\right|^{2}+4-8=4a^{2}$

$\therefore\left|MN\right|^{2}=4a^{2}+4$

$\therefore 4c^{2}-8=4a^{2}+4$

$\therefore c^{2}=a^{2}+3$

$\therefore e^{2}+\dfrac{a^{2}}{4}=\dfrac{c^{2}}{a^{2}}+\dfrac{a^{2}}{4}=\dfrac{a^{2}+3}{a^{2}}+\dfrac{a^{2}}{4}=1+\dfrac{3}{a^{2}}+\dfrac{a^{2}}{4}\geq 1+\sqrt{3}$