Question

9．將曲線$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1按φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$變換后的曲線的參數(shù)方程為（θ為參數(shù)）（　�。�

• A． $\left\{\begin{array}{l}{x=3cosθ}\\{y=2sinθ}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=\sqrt{3}cosθ}\\{y=\sqrt{2}sinθ}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=\frac{1}{3}cosθ}\\{y=\frac{1}{2}sinθ}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=\frac{\sqrt{3}}{3}cosθ}\\{y=\frac{\sqrt{2}}{2}sinθ}\end{array}\right.$

Answer 1

分析 由變換φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$，可得：$\left\{\begin{array}{l}{x=3{x}^{′}}\\{y=2{y}^{′}}\end{array}\right.$，代入曲線$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1，再利用同角三角函數(shù)的基本關(guān)系式即可得出．

解答 解：由變換φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$，可得：$\left\{\begin{array}{l}{x=3{x}^{′}}\\{y=2{y}^{′}}\end{array}\right.$，代入曲線$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1可得：3（x′）²+2（y′）²=1，
即為：3x²+2y²=1，令$\left\{\begin{array}{l}{x=\frac{\sqrt{3}}{3}cosθ}\\{y=\frac{\sqrt{2}}{2}sinθ}\end{array}\right.$（θ為參數(shù)）即可得出參數(shù)方程．
故選：D．

點(diǎn)評(píng) 本題考查了橢圓的參數(shù)方程、坐標(biāo)變換、同角三角函數(shù)基本關(guān)系式，考查了推理能力與計(jì)算能力，屬于中檔題．