Question

8．在平面直角坐标中，经伸缩变换后曲线x²+y²=16变换为椭圆x′²+$\frac{y'}{16}^2}$=1，此伸缩变换公式是（　　）

• 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 设伸缩变换为$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，代入x′²+$\frac{y'}{16}^2}$=1，与x²+y²=16比较，即可得出结论．

解答 解：设伸缩变换为$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，代入x′²+$\frac{y'}{16}^2}$=1      
得到（λx）²+$\frac{1}{16}$（μy）²=1，即16λ²x²+μ²y²=16  ①
将①式与x²+y²=16变比较，得λ=$\frac{1}{4}$，μ=1
故所求的伸缩变换为$\left\{\begin{array}{l}{x=4x′}\\{y=y′}\end{array}\right.$．
故选：B．

点评 本题考查了圆变换为椭圆的伸缩变换，考查了变形能力与计算能力，属于中档题．