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=1 | B. | 9x2+25y2=1 | C. | 25x+9y=1 | D. | $\frac{{x}^{2}}{25}$+$\frac{{y}^{2}}{9}$=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.
练习册系列答案
相关题目
10.将一颗正方体型骰子投掷2次,向上的点数之和是6的概率( )
A. | $\frac{1}{12}$ | B. | $\frac{1}{9}$ | C. | $\frac{5}{36}$ | D. | $\frac{1}{36}$ |
11.已知数列-1,x,y,z,-3为等比数列,则xyz=( )
A. | 9 | B. | ±9 | C. | $-3\sqrt{3}$ | D. | $±3\sqrt{3}$ |
5.已知四边形ABCD为平行四边形,A(-1,2),B(0,0),C(1,7),则点D的坐标是( )
A. | (-9,9) | B. | (-9,0) | C. | (0,9) | D. | (0,-9) |