Question

15．給定矩陣A=$[\begin{array}{l}{1}&{2}\\{2}&{3}\end{array}]$，B=$[\begin{array}{l}{-\frac{3}{2}}&{2}\\{1}&{-1}\end{array}]$，設(shè)橢圓$\frac{{x}^{2}}{4}$+y²=1在矩陣AB對應(yīng)的變換下得到曲線F，求F的面積．

Answer 1

分析 求出AB，可得橢圓$\frac{{x}^{2}}{4}$+y²=1在矩陣AB對應(yīng)的變換下得到曲線F，即可求F的面積．

解答 解：由已知得$AB=[\begin{array}{l}1\\ 2\end{array}\right.\;\;\;\left.\begin{array}{l}2\\ 3\end{array}][\begin{array}{l}-\frac{3}{2}\;\;\;\;2\\ \;\;1\;\;\;\;\;-1\end{array}]=[\begin{array}{l}\frac{1}{2}\;\;\;\;0\\ \;0\;\;\;\;1\end{array}]$，
設(shè)P（x₀，y₀）為橢圓上任意一點，點M在矩陣AB對應(yīng)的變換下變?yōu)辄c$P'（{x_0}^′，{y_0}^′）$，
則有$[\begin{array}{l}\frac{1}{2}\;\;\;\;0\\ \;0\;\;\;\;1\end{array}][\begin{array}{l}{x_0}\\{y_0}\end{array}]=[\begin{array}{l}{x_0}^′\\{y_0}^′\end{array}]$，即$\left\{\begin{array}{l}\frac{1}{2}{x_0}={x_0}^′\\{y_0}={y_0}^′\end{array}\right.$，所以$\left\{\begin{array}{l}{x_0}=2{x_0}^′\\{y_0}={y_0}^′\end{array}\right.$，
又點P在橢圓上，故$\frac{{{x_0}^2}}{4}+{y_0}^2=1$，從而${（{x_0}^′）^2}+{（{y_0}^′）^2}=1$，
故曲線F的方程為x²+y²=1，其面積為π．

點評 本題考查矩陣與變換，考查圓的方程，屬于基礎(chǔ)題．