Question

9．将曲线$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1按φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$变换后的曲线的参数方程为（θ为参数）（　　）

• 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 由变换φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$，可得：$\left\{\begin{array}{l}{x=3{x}^{′}}\\{y=2{y}^{′}}\end{array}\right.$，代入曲线$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1，再利用同角三角函数的基本关系式即可得出．

解答 解：由变换φ：$\left\{\begin{array}{l}{x′=\frac{1}{3}x}\\{y′=\frac{1}{2}y}\end{array}\right.$，可得：$\left\{\begin{array}{l}{x=3{x}^{′}}\\{y=2{y}^{′}}\end{array}\right.$，代入曲线$\frac{{x}^{2}}{3}$+$\frac{{y}^{2}}{2}$=1可得：3（x′）²+2（y′）²=1，
即为：3x²+2y²=1，令$\left\{\begin{array}{l}{x=\frac{\sqrt{3}}{3}cosθ}\\{y=\frac{\sqrt{2}}{2}sinθ}\end{array}\right.$（θ为参数）即可得出参数方程．
故选：D．

点评 本题考查了椭圆的参数方程、坐标变换、同角三角函数基本关系式，考查了推理能力与计算能力，属于中档题．