25~26高三下·山东临沂一模·第8题

A

设$|AB| = m$，则$|AF_1| = |AB| = m$.

由双曲线的定义，可得$|AF_1| – |AF_2| = 2a$，所以$|AF_2| = m – 2a$.

又$|BF_2| = |AB| – |AF_2| = m – (m – 2a) = 2a$.

因为$|BF_1| – |BF_2| = 2a$，所以$|BF_1| = |BF_2| + 2a = 2a + 2a = 4a$.

在$\triangle ABF_1$中，由余弦定理：
$
|BF_1|^2 = |AB|^2 + |AF_1|^2 – 2|AB|\cdot|AF_1|\cos\angle BAF_1
$

代入得：
$
(4a)^2 = m^2 + m^2 – 2 \cdot m^2 \cdot \dfrac{1}{9}
$

$
16a^2 = 2m^2 – \dfrac{2m^2}{9} = \dfrac{16m^2}{9}
$

即$m^2 = 9a^2$，所以$m = 3a$.

则$|AF_1| = 3a$，$|AF_2| = 3a – 2a = a$.

在$\triangle AF_1F_2$中，$\angle F_1AF_2 = \angle BAF_1$，由余弦定理：
$
|F_1F_2|^2 = |AF_1|^2 + |AF_2|^2 – 2|AF_1|\cdot|AF_2|\cos\angle F_1AF_2
$

代入得：
$
(2c)^2 = (3a)^2 + a^2 – 2 \cdot 3a \cdot a \cdot \dfrac{1}{9}
$

$
4c^2 = 9a^2 + a^2 – \dfrac{6a^2}{9} = 10a^2 – \dfrac{2a^2}{3} = \dfrac{28a^2}{3}
$

化简得$c^2 = \dfrac{7}{3}a^2$，故$\dfrac{c^2}{a^2} = \dfrac{7}{3}$.

所以双曲线的离心率$e = \dfrac{c}{a} = \sqrt{\dfrac{7}{3}} = \dfrac{\sqrt{21}}{3}$.