已知双曲线\(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) \((a>0,b>0)\)的左,右焦点分别为\(F_1(-c,0)\),\(F_2(c,0)\),直线\(x=m\)交双曲线于\(A,B\)两点,且\(AF_2\perp BF_1\),\(AF_2\)与\(y\)轴的交点为\(C(0,-3b)\),则双曲线的离心率为\((\qquad)\)
\(\mathrm{A}.\dfrac{3\sqrt{5}}{5}\) \(\qquad\mathrm{B}.3\) \(\qquad\mathrm{C}.\sqrt{3}+1\) \(\qquad\mathrm{D}.\sqrt{2}\)
解析:
如图所示,若设$\angle AF_2M=\theta$,则$$ \angle AF_1M=\angle BF_1M=\angle MAF_2=\dfrac{\pi}{2}-\theta.$$于是在$\triangle AF_1F_2$中,$\angle F_1AF_2=2\theta-\dfrac{\pi}{2}$.从而由正弦定理可得$$ \dfrac{|F_1F_2|}{\sin\left(2\theta-\dfrac{\pi}{2}\right)}=\dfrac{|AF_1|}{\sin\left(\pi-\theta\right)}=\dfrac{|AF_2|}{\sin\left(\dfrac{\pi}{2}-\theta\right)}=\dfrac{|AF_1|-|AF_2|}{\sin\theta-\cos\theta}.$$再结合双曲线的第一定义可知有$$\dfrac{2c}{\cos 2\theta}=\dfrac{2a}{\cos\theta-\sin\theta}.$$从而离心率$e=\cos\theta+\sin\theta$.两边平方可得$$ \dfrac{c^2}{c^2-b^2}=1+\dfrac{\tan\theta}{1+\tan^2\theta}=1+\dfrac{\dfrac{3b}{c}}{1+\left(\dfrac{3b}{c}\right)^2}.$$若设$t=\dfrac{c}{b}$,则将上式整理可得$$ 6x^3-x^2-6x-9=0.$$ 易估计$2>x>1$,再结合有理根定理试根,只有$\dfrac{3}{2}$满足,因此$x=\dfrac{3}{2}$,从而所求离心率为$\dfrac{3\sqrt{5}}{5}$.
