Question

1．在平面直角坐標系xOy內(nèi)，變換D₁．將每個點（x，y）沿著與x軸平行的方向平移2y個單位變成點P′．變換D₂將點（x，y）變?yōu)椋▁′，y′），其坐標變換公式為$\left\{\begin{array}{l}x'=x\\ y'=2y.\end{array}\right.$
（Ⅰ）寫出D1的坐標變換公式及D_l、D₂所對應的二階矩陣A、B；
（Ⅱ）求曲線C：x²-4y²=1依次經(jīng)過D_l和D₂變換作用后的曲線C′的方程．

Answer 1

分析 （Ⅰ）根據(jù)坐標變換，$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{x′}\\{y′}\end{array}）$，$（\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}）$$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{x′}\\{y′}\end{array}）$代入求得矩陣A和B；
（Ⅱ）根據(jù)矩陣的乘法公式BA=$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$×$（\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}）$=$（\begin{array}{l}{1}&{2}\\{0}&{2}\end{array}）$，$\left\{\begin{array}{l}{x′=x+2y}\\{y′=2y}\end{array}\right.$，代入求得$\left\{\begin{array}{l}{x=x′-y′}\\{y=\frac{1}{2}y′}\end{array}\right.$，代入曲線C，求得曲線C′的方程．

解答 解：（Ⅰ）D₁坐標變換公式為$\left\{\begin{array}{l}{x′=x+2y}\\{y′=y}\end{array}\right.$，
∴$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{x′}\\{y′}\end{array}）$
∴A=$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$，
由$\left\{\begin{array}{l}x'=x\\ y'=2y.\end{array}\right.$，
$（\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}）$$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{x′}\\{y′}\end{array}）$
∴B=$（\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}）$，
（Ⅱ）由題意可知經(jīng)過D_l和D₂變換所對應的矩陣為：
BA=$（\begin{array}{l}{1}&{2}\\{0}&{1}\end{array}）$×$（\begin{array}{l}{1}&{0}\\{0}&{2}\end{array}）$=$（\begin{array}{l}{1}&{2}\\{0}&{2}\end{array}）$，
坐標變換公式為：
$\left\{\begin{array}{l}{x′=x+2y}\\{y′=2y}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x=x′-y′}\\{y=\frac{1}{2}y′}\end{array}\right.$，
代入曲線C得x′²-2x′y′=1，即x²-2xy=1．

點評 本題考查矩陣與變換、曲線在矩陣變換下的曲線的方程，考查矩陣的乘法，屬于中檔題．