Question

20．曲线x²+y²=1经过伸缩变换$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$后，变成的曲线方程是（　　）

• A． 25x²+9y²=1
• B． 9x²+25y²=1
• C． 25x+9y=1
• D． $\frac{{x}^{2}}{25}$+$\frac{{y}^{2}}{9}$=1

Answer 1

分析 由伸缩变换$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$，化为$\left\{\begin{array}{l}{x=5{x}^{′}}\\{y=3{y}^{′}}\end{array}\right.$，代入曲线x²+y²=1即可得出．

解答 解：由伸缩变换$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$，化为$\left\{\begin{array}{l}{x=5{x}^{′}}\\{y=3{y}^{′}}\end{array}\right.$，代入曲线x²+y²=1可得25（x′）²+9（y′）²=1，
故选：A．

点评 本题考查了伸缩变换及其化简能力，属于基础题．