Question

12．在平面直角坐標(biāo)系中，經(jīng)伸縮變換后曲線方程x²+y²=4變換為橢圓方程x′²+$\frac{y{′}^{2}}{4}$=1，此伸縮變換公式是（　�。�

• A． $\left\{\begin{array}{l}{x=\frac{1}{2}x′}\\{x=y′}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2x′}\\{y=y′}\end{array}\right.$
• C． $\left\{\begin{array}{l}{y=4x′}\\{y=y′}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2x′}\\{y=4y′}\end{array}\right.$

Answer 1

分析 經(jīng)伸縮變換后曲線方程x²+y²=4即$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{4}$=1，變換為橢圓方程x′²+$\frac{y{′}^{2}}{4}$=1，可得變換公式$\left\{\begin{array}{l}{{x}^{′}=\frac{x}{2}}\\{{y}^{′}=y}\end{array}\right.$，即可得出．

解答 解：∵經(jīng)伸縮變換后曲線方程x²+y²=4即$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{4}$=1，變換為橢圓方程x′²+$\frac{y{′}^{2}}{4}$=1，∴$\left\{\begin{array}{l}{{x}^{′}=\frac{x}{2}}\\{{y}^{′}=y}\end{array}\right.$，即$\left\{\begin{array}{l}{x=2{x}^{′}}\\{y={y}^{′}}\end{array}\right.$，
故選：B．

點(diǎn)評(píng) 本題考查了坐標(biāo)變換，考查了推理能力與計(jì)算能力，屬于基礎(chǔ)題．