题目内容
3.设F2(c,0)(c>0)是双曲线Γ:$\frac{{x}^{2}}{{a}^{2}}$-$\frac{{y}^{2}}{{b}^{2}}$=1(a>0,b>0)的右焦点,M是双曲线左支上的一点,线段MF2与圆x2+y2-$\frac{2c}{3}$x+$\frac{{a}^{2}}{9}$=0相切于D,且|MF2|=3|DF2|,则双曲线Γ的离心率为( )A. | $\sqrt{3}$ | B. | 2 | C. | $\sqrt{5}$ | D. | $\frac{\sqrt{5}}{2}$ |
分析 由圆的方程求出圆心坐标,设出D的坐标,由题意列式求出D的坐标,结合|MF2|=3|DF2|求得M的坐标,再把M的坐标代入双曲线方程求得答案.
解答 解:由x2+y2-$\frac{2c}{3}$x+$\frac{{a}^{2}}{9}$=0,得$(x-\frac{c}{3})^{2}+{y}^{2}=\frac{{b}^{2}}{9}$,
则该圆的圆心坐标为($\frac{c}{3}$,0),半径为$\frac{b}{3}$.
设切点D(x0,y0)(y0>0),
则$\left\{\begin{array}{l}{{{x}_{0}}^{2}+{{y}_{0}}^{2}-\frac{2c}{3}{x}_{0}+\frac{{a}^{2}}{9}=0}\\{({x}_{0}-c,{y}_{0})•({x}_{0}-\frac{c}{3},{y}_{0})=0}\end{array}\right.$,
∴$\left\{\begin{array}{l}{{{x}_{0}}^{2}+{{y}_{0}}^{2}-\frac{2c}{3}{x}_{0}+\frac{{a}^{2}}{9}=0}\\{{{x}_{0}}^{2}-\frac{4c}{3}{x}_{0}+{{y}_{0}}^{2}+\frac{{c}^{2}}{3}=0}\end{array}\right.$,解得:$\left\{\begin{array}{l}{{x}_{0}=\frac{3{c}^{2}-{a}^{2}}{6c}}\\{{y}_{0}=\frac{b\sqrt{3{c}^{2}+{a}^{2}}}{6c}}\end{array}\right.$.
∴D($\frac{3{c}^{2}-{a}^{2}}{6c},\frac{b\sqrt{3{c}^{2}+{a}^{2}}}{6c}$),
由|MF2|=3|DF2|,得M($-\frac{{a}^{2}+{c}^{2}}{2c}$,$\frac{b\sqrt{3{c}^{2}+{a}^{2}}}{2c}$),
代入$\frac{{x}^{2}}{{a}^{2}}$-$\frac{{y}^{2}}{{b}^{2}}$=1,得:
$\frac{({a}^{2}+{c}^{2})^{2}}{4{a}^{2}{c}^{2}}-\frac{3{c}^{2}+{a}^{2}}{4{c}^{2}}=1$,整理得,$\frac{{c}^{2}}{{a}^{2}}=5$,∴$e=\sqrt{5}$.
故选:C.
点评 本题考查了双曲线的简单几何性质,考查了圆与圆锥曲线间的关系,考查了学生的计算能力,是中档题.