Question

8．在平面直角坐標(biāo)中，經(jīng)伸縮變換后曲線x²+y²=16變換為橢圓x′²+$\frac{y'}{16}^2}$=1，此伸縮變換公式是（　�。�

• A． $\left\{{\begin{array}{l}{x=\frac{1}{4}x'}\\{y=y'}\end{array}}\right.$
• B． $\left\{{\begin{array}{l}{x=4x'}\\{y=y'}\end{array}}\right.$
• C． $\left\{{\begin{array}{l}{x=2x'}\\{y=y'}\end{array}}\right.$
• D． $\left\{{\begin{array}{l}{x=4x'}\\{y=8y'}\end{array}}\right.$

Answer 1

分析 設(shè)伸縮變換為$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，代入x′²+$\frac{y'}{16}^2}$=1，與x²+y²=16比較，即可得出結(jié)論．

解答 解：設(shè)伸縮變換為$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，代入x′²+$\frac{y'}{16}^2}$=1      
得到（λx）²+$\frac{1}{16}$（μy）²=1，即16λ²x²+μ²y²=16  ①
將①式與x²+y²=16變比較，得λ=$\frac{1}{4}$，μ=1
故所求的伸縮變換為$\left\{\begin{array}{l}{x=4x′}\\{y=y′}\end{array}\right.$．
故選：B．

點評 本題考查了圓變換為橢圓的伸縮變換，考查了變形能力與計算能力，屬于中檔題．