题目内容

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

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