浙江 2016 · 数学 q07

题面

7．（5分）（2016•浙江）已知椭圆$\mathrm{C_{1}} \colon \mathrm{\frac{x}{m^{2}}} \mathrm{+ y}^{2} \mathrm{= 1 \ ( m > 1 )}$与双曲线$\mathrm{C}_{2} \colon \mathrm{\frac{{x}^{2}}{{n}^{2}} - y^{2} = 1 \mathrm{( n > 0 )}}$的焦点重合，$e_{1} , e_{2}$分别为$\mathrm{C}_{1} , \mathrm{C}_{2}$的离心率，则（ ）

答案

A

解析

分析

根据椭圆和双曲线有相同的焦点，得到$c^{2} = m^{2} - 1 = n^{2} + 1$，即$m^{2} \cdot n^{2}{=} 2$，进行判断，能得$m > n$，求出两个离心率，先平方进行化简进行判断即可

解答

解：∵椭圆$\mathrm{C_{1}} \colon \mathrm{\frac{x}{m^{2}}} \mathrm{+ y}^{2} \mathrm{= 1 \ ( m > 1 )}$与双曲线$\mathrm{C}_{2} \colon \mathrm{\frac{{x}}{\mathrm{}^{2}} - y^{2} = 1 \mathrm{( n} > 0 )}$）的焦点重合，

∴满足$c^{2} = m^{2} - 1 = n^{2} + 1$

即$\mathrm{m}^{2} - \mathrm{n}^{2}{=} 2 {>} 0 , {\div} \mathrm{m}^{2}{>} \mathrm{n}^{2}$，则$\mathfrak{m} > \mathfrak{n}$，排除C，D

则$\mathrm{c}^{2} \mathrm{= m}^{2} - 1 {\mathrm{< m}}^{2} , \mathrm{c}^{2} \mathrm{= n}^{2} \mathrm{+} 1 \mathrm{> n}^{2}$

则$c {<} m . c {>} n$

$\mathrm{e}_{1} = \frac{\mathrm{c}}{\pi}, \quad \mathrm{e}_{2} = \frac{\mathrm{c}}{\mathrm{n}},$

则$e_{1} \bullet e_{2} = {\frac{c_{\bullet} \circ \underline{{c}}}{\pi{n} \ \Pi \Pi}} ,$

$\begin{array}{l} \text{则} (\mathrm{e}_{1} \bullet \mathrm{e}_{2})^{2} = (\frac{\mathrm{C}}{\pi})^{2} \bullet (\frac{\mathrm{C}}{\mathrm{n}}) \\ 2 = \frac{\mathrm{c}^{2}}{\mathrm{m}^{2}} \cdot \frac{\mathrm{c}^{2}}{\mathrm{n}^{2}} = \frac{(\mathrm{m}^{2} - 1) (\mathrm{n}^{2} + 1)}{\mathrm{m}^{2} \mathrm{n}^{2}} = \frac{\mathrm{m}^{2} \mathrm{n}^{2} + (\mathrm{m}^{2} - \mathrm{n}^{2}) - 1}{\mathrm{m}^{2} \mathrm{n}^{2}} = 1 + \frac{\mathrm{m}^{2} - \mathrm{n}^{2} - 1}{\mathrm{m}^{2} \mathrm{n}^{2}} = 1 + \frac{2 - 1}{\mathrm{m}^{2} \mathrm{n}^{2}} \\ 1 + \frac{1}{\mathrm{m}^{2} \mathrm{n}^{2}} > 1, \\ \therefore \mathrm{e}_{1} \mathrm{e}_{2} > 1, \\ \text{故选：A.} \end{array}$