Question

10．變換T₁是逆時(shí)針旋轉(zhuǎn)$\frac{π}{2}$角的旋轉(zhuǎn)變換，對(duì)應(yīng)的變換矩陣是M₁；變換T₂對(duì)應(yīng)的變換矩陣是M₂=$[\begin{array}{l}{1}&{1}\\{0}&{1}\end{array}]$．
（1）點(diǎn)P（2，1）經(jīng)過變換T₁得到點(diǎn)P′，求P′的坐標(biāo)；
（2）求曲線y=x²先經(jīng)過變換T₁，再經(jīng)過變換T₂所得曲線的方程．

Answer 1

分析 （1）變換T₁對(duì)應(yīng)的變換矩陣M₁=$[\begin{array}{l}{cos\frac{π}{2}}&{-sin\frac{π}{2}}\\{sin\frac{π}{2}}&{cos\frac{π}{2}}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$，M₁$[\begin{array}{l}{2}\\{1}\end{array}]$=$[\begin{array}{l}{-1}\\{2}\end{array}]$，即可求得點(diǎn)P在T₁作用下的點(diǎn)P′的坐標(biāo)；
（2）M=M₂•M₁=$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$，由$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，求得$\left\{\begin{array}{l}{{x}_{0}=y}\\{{y}_{0}=y-x}\end{array}\right.$，代入y=x²，即可求得經(jīng)過變換T₂所得曲線的方程．

解答 解：（1）T₁是逆時(shí)針旋轉(zhuǎn)$\frac{π}{2}$角的旋轉(zhuǎn)變換，M₁=$[\begin{array}{l}{cos\frac{π}{2}}&{-sin\frac{π}{2}}\\{sin\frac{π}{2}}&{cos\frac{π}{2}}\end{array}]$=$[\begin{array}{l}{0}&{-1}\\{1}&{0}\end{array}]$，
M₁$[\begin{array}{l}{2}\\{1}\end{array}]$=$[\begin{array}{l}{-1}\\{2}\end{array}]$，
所以點(diǎn)P在T₁作用下的點(diǎn)P′的坐標(biāo)是（-1，2）；
（2）M=M₂•M₁=$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$，
設(shè)$[\begin{array}{l}{x}\\{y}\end{array}]$是變換后圖象上任一點(diǎn)，與之對(duì)應(yīng)的變換前的點(diǎn)是$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$，
則M$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，$[\begin{array}{l}{1}&{-1}\\{1}&{0}\end{array}]$$[\begin{array}{l}{{x}_{0}}\\{{y}_{0}}\end{array}]$=$[\begin{array}{l}{x}\\{y}\end{array}]$，
也就是$\left\{\begin{array}{l}{{x}_{0}-{y}_{0}=x}\\{{x}_{0}=y}\end{array}\right.$，即$\left\{\begin{array}{l}{{x}_{0}=y}\\{{y}_{0}=y-x}\end{array}\right.$，
所以所求的曲線方程為y-x=y²．

點(diǎn)評(píng) 本題考查矩陣的變換，考查矩陣的乘法，考查點(diǎn)在變換下點(diǎn)的坐標(biāo)的求法，屬于中檔題．