Question

20．曲線x²+y²=1經(jīng)過伸縮變換$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$后，變成的曲線方程是（　�。�

• 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 由伸縮變換$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$，化為$\left\{\begin{array}{l}{x=5{x}^{′}}\\{y=3{y}^{′}}\end{array}\right.$，代入曲線x²+y²=1即可得出．

解答 解：由伸縮變換$\left\{\begin{array}{l}{x′=\frac{1}{5}x}\\{y′=\frac{1}{3}y}\end{array}\right.$，化為$\left\{\begin{array}{l}{x=5{x}^{′}}\\{y=3{y}^{′}}\end{array}\right.$，代入曲線x²+y²=1可得25（x′）²+9（y′）²=1，
故選：A．

點評 本題考查了伸縮變換及其化簡能力，屬于基礎(chǔ)題．