浙江 2013 · 数学 q09

题面

9．（5分）（2013•浙江）如图F₁、F₂是椭圆C₁：$\frac{x^{2}}{4} + y^{2} = 1$与双曲线C₂的公共焦点 A、B 分别是 C₁、C₂在第二、四象限的公共点，若四边形$\mathrm{A F}_{1} \mathrm{B F}_{2}$为矩形，则$\mathrm{C}_{2}$的离心率是（ ）

A．$\sqrt{2}$ B．$\sqrt{3}$ C．$\frac{3}{2}$ D．$\frac{\sqrt{6}}{2}$

答案

D

解析

分析

不妨设$| \mathrm{A F}_{1} | = x , | \mathrm{A F}_{2} | = y$，依题意${\left\{\begin{array}{l l}{x + y = 4} \\ {x^{2} + y^{2} = 1 2} \end{array} \right.}$，解此方程组可求得x，y的值，利用双曲线的定义及性质即可求得$\mathrm{C}_{2}$的离心率

解答

解：设$| \mathrm{A F}_{1} | = x , | \mathrm{A F}_{2} | = y$，∵点A为椭圆C₁：$\frac{x^{2}}{4} + y^{2} = 1$上的点，

$\therefore 2 a = 4 , \ b = 1 , \ c = \sqrt{3}$；

$\therefore | \mathrm{A F}_{1} | + | \mathrm{A F}_{2} | = 2 \mathrm{a} = 4$，即$x + y = 4$；①

又四边形$\mathrm{A F}_{1} \mathrm{B F}_{2}$为矩形，

$\therefore | \mathrm{A F}_{1} |^{2} + | \mathrm{A F}_{2} |^{2} = | \mathrm{F}_{1} \mathrm{F}_{2} |^{2}$，即$x^{2} + y^{2} = ( 2 c )^{2} = ( 2 \sqrt{3} )^{2} = 1 2$，②

由$\textcircled{1} \textcircled{2}$得：${\left\{\begin{array}{l l}{x + y = 4} \\ {x^{2} + y^{2} = 1 2} \end{array} \right.}$，解得$\mathrm{x} = 2 - \sqrt{2} , \mathrm{y} = 2 + \sqrt{2}$，设双曲线$\mathrm{C}_{2}$的实轴长为2a，焦距为$2 \mathrm{c}$，

则$2 \mathrm{a} = , | \mathrm{A F}_{2} | - | \mathrm{A F}_{1} | = \mathrm{y} - \mathrm{x} = 2 \sqrt{2} , 2 \mathrm{c} = 2 \sqrt{2^{2} - 1^{2}} = 2 \sqrt{3}$，

∴双曲线C₂的离心率$\mathrm{e} = \frac{c}{a} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}$

故选D