Question

5．可以将椭圆$\frac{{x}^{2}}{10}$+$\frac{{y}^{2}}{8}$=1变为圆x²+y²=4的伸缩变换为（　　）

• A． $\left\{\begin{array}{l}{x′=\frac{2}{5}x}\\{y′=\frac{\sqrt{2}}{2}y}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x′=\frac{\sqrt{10}}{2}x}\\{y′=\sqrt{2}y}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x′=\frac{\sqrt{2}}{2}x}\\{y′=\frac{\sqrt{10}}{5}y}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x′=\frac{\sqrt{10}}{5}x}\\{y′=\frac{\sqrt{2}}{2}y}\end{array}\right.$

Answer 1

分析 令$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$代入，化简代入椭圆方程化简整理即可得出．

解答 解：由圆x²+y²=4化为$（\frac{x}{2}）^{2}+（\frac{y}{2}）^{2}$=1，
令$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$代入椭圆方程可得$\frac{（{x}^{′}）^{2}}{4}+\frac{（{y}^{′}）^{2}}{4}$=1，即（x′）²+（y′）²=4，
由$\left\{\begin{array}{l}{x=\frac{\sqrt{10}}{2}{x}^{′}}\\{y=\sqrt{2}{y}^{′}}\end{array}\right.$化为$\left\{\begin{array}{l}{{x}^{′}=\frac{\sqrt{10}}{5}x}\\{{y}^{′}=\frac{\sqrt{2}}{2}y}\end{array}\right.$．
故选：D．

点评 本题考查了椭圆化为圆的变换公式，考查了计算能力，属于基础题．