Question

15．给定矩阵A=$[\begin{array}{l}{1}&{2}\\{2}&{3}\end{array}]$，B=$[\begin{array}{l}{-\frac{3}{2}}&{2}\\{1}&{-1}\end{array}]$，设椭圆$\frac{{x}^{2}}{4}$+y²=1在矩阵AB对应的变换下得到曲线F，求F的面积．

Answer 1

分析 求出AB，可得椭圆$\frac{{x}^{2}}{4}$+y²=1在矩阵AB对应的变换下得到曲线F，即可求F的面积．

解答 解：由已知得$AB=[\begin{array}{l}1\\ 2\end{array}\right.\;\;\;\left.\begin{array}{l}2\\ 3\end{array}][\begin{array}{l}-\frac{3}{2}\;\;\;\;2\\ \;\;1\;\;\;\;\;-1\end{array}]=[\begin{array}{l}\frac{1}{2}\;\;\;\;0\\ \;0\;\;\;\;1\end{array}]$，
设P（x₀，y₀）为椭圆上任意一点，点M在矩阵AB对应的变换下变为点$P'（{x_0}^′，{y_0}^′）$，
则有$[\begin{array}{l}\frac{1}{2}\;\;\;\;0\\ \;0\;\;\;\;1\end{array}][\begin{array}{l}{x_0}\\{y_0}\end{array}]=[\begin{array}{l}{x_0}^′\\{y_0}^′\end{array}]$，即$\left\{\begin{array}{l}\frac{1}{2}{x_0}={x_0}^′\\{y_0}={y_0}^′\end{array}\right.$，所以$\left\{\begin{array}{l}{x_0}=2{x_0}^′\\{y_0}={y_0}^′\end{array}\right.$，
又点P在椭圆上，故$\frac{{{x_0}^2}}{4}+{y_0}^2=1$，从而${（{x_0}^′）^2}+{（{y_0}^′）^2}=1$，
故曲线F的方程为x²+y²=1，其面积为π．

点评 本题考查矩阵与变换，考查圆的方程，属于基础题．