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\(\dfrac{3\sqrt{5}}{5}\).每日一题_191018
原文：https://www.cnblogs.com/Math521/p/11681894.html
