PApB分别交椭圆另外的点MN

16. 已知 $A$，$B$ 是椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\ (a>b>0)$ 的左右顶点，$P$ 是双曲线 $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ 在第一象限上的一点，直线 $PA$，$PB$ 分别交椭圆于另外的点 $M$，$N$。若直线 $MN$ 过椭圆的右焦点 $F$，且 $\tan\angle AMN=3$，则椭圆的离心率为__。

解： 设 $P\left(x_{1},y_{1}\right)$，$M\left(x_{2},y_{2}\right)$

$\therefore \dfrac{x_{1}^{2}}{a^{2}}-\dfrac{y_{1}^{2}}{b^{2}}=1$，$\dfrac{x_{2}^{2}}{a^{2}}+\dfrac{y_{2}^{2}}{b^{2}}=1$

$\therefore y_{1}^{2}=\dfrac{b^{2}}{a^{2}}\left(x_{1}^{2}-a^{2}\right)$，$y_{2}^{2}=-\dfrac{b^{2}}{a^{2}}\left(x_{2}^{2}-a^{2}\right)$

$\therefore K_{PA}\times K_{PB}=\dfrac{y_{1}}{x_{1}+a}\times\dfrac{y_{1}}{x_{1}-a}=\dfrac{y_{1}^{2}}{x_{1}^{2}-a^{2}}=\dfrac{b^{2}}{a^{2}}$

$K_{MA}\times K_{MB}=\dfrac{y_{2}}{x_{2}+a}\times\dfrac{y_{2}}{x_{2}-a}=\dfrac{y_{2}^{2}}{x_{2}^{2}-a^{2}}=-\dfrac{b^{2}}{a^{2}}$

$\therefore K_{PB}=-K_{MB}$

$\therefore MB$ 与 $PB$ 关于 $x$ 轴对称

$\therefore$ 点 $M$ 与点 $N$ 关于 $x$ 轴对称

$\therefore MN\perp x$ 轴

$\therefore x_{2}=c$，$y_{2}=\dfrac{b^{2}}{a}$

$\therefore \tan\angle AMN=\dfrac{a+c}{\dfrac{b^{2}}{a}}=3$

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

$\therefore e=\dfrac{2}{3}$