Question

1．已知曲線C：x²-xy+y²=3，矩陣$M=（{\begin{array}{l}{\frac{{\sqrt{2}}}{2}}&{\frac{{\sqrt{2}}}{2}}\\{-\frac{{\sqrt{2}}}{2}}&{\frac{{\sqrt{2}}}{2}}\end{array}}）$，且曲線C在矩陣M對(duì)應(yīng)的變換的作用下得到曲線C′．
（Ⅰ）求曲線C′的方程；
（Ⅱ）求曲線C的離心率以及焦點(diǎn)坐標(biāo)．

Answer 1

分析 （Ⅰ）利用變換的定義直接計(jì)算即可；
（Ⅱ）通過（I）可知曲線C′的離心率及焦點(diǎn)坐標(biāo)，利用逆矩陣可得結(jié)論．

解答 解：（Ⅰ）設(shè)曲線C′上的任一點(diǎn)P（x，y）在矩陣M對(duì)應(yīng)的變換作用下變?yōu)辄c(diǎn)P′（x′，y′），
則有$（\begin{array}{l}{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\\{-\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}）$$（\begin{array}{l}{x}\\{y}\end{array}）$=$（\begin{array}{l}{x′}\\{y′}\end{array}）$，即$\left\{\begin{array}{l}{x′=\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y}\\{y′=-\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y}\end{array}\right.$，
所以$\left\{\begin{array}{l}{x=\frac{\sqrt{2}}{2}x′-\frac{\sqrt{2}}{2}y′}\\{y=\frac{\sqrt{2}}{2}x′+\frac{\sqrt{2}}{2}y′}\end{array}\right.$，
將上式代入曲線C的方程，整理得$\frac{x{′}^{2}}{6}+\frac{y{′}^{2}}{2}=1$，
∴曲線C′的方程為$\frac{{x}^{2}}{6}+\frac{{y}^{2}}{2}=1$；
（Ⅱ）由（I）可知曲線C′的離心率$e=\frac{\sqrt{6}}{3}$，焦點(diǎn)坐標(biāo)為F₁′（-2，0），F(xiàn)₂′（2，0），
∵M(jìn)=$（\begin{array}{l}{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\\{-\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}）$對(duì)應(yīng)的變換是順時(shí)針旋轉(zhuǎn)$\frac{π}{4}$的旋轉(zhuǎn)變換，
∴M^-1=$（\begin{array}{l}{\frac{\sqrt{2}}{2}}&{-\frac{\sqrt{2}}{2}}\\{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}）$對(duì)應(yīng)的變換是逆時(shí)針旋轉(zhuǎn)$\frac{π}{4}$的旋轉(zhuǎn)變換，
∴曲線C的離心率為$\frac{\sqrt{6}}{3}$，且曲線C的焦點(diǎn)是曲線C′的焦點(diǎn)經(jīng)過逆時(shí)針旋轉(zhuǎn)$\frac{π}{4}$得到．
∵$（\begin{array}{l}{\frac{\sqrt{2}}{2}}&{-\frac{\sqrt{2}}{2}}\\{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}）$$（\begin{array}{l}{-2}\\{0}\end{array}）$=$（\begin{array}{l}{-\sqrt{2}}\\{-\sqrt{2}}\end{array}）$$（\begin{array}{l}{\frac{\sqrt{2}}{2}}&{-\frac{\sqrt{2}}{2}}\\{\frac{\sqrt{2}}{2}}&{\frac{\sqrt{2}}{2}}\end{array}）$$（\begin{array}{l}{2}\\{0}\end{array}）$=$（\begin{array}{l}{\sqrt{2}}\\{\sqrt{2}}\end{array}）$，
∴曲線C的焦點(diǎn)為（$-\sqrt{2}$，$-\sqrt{2}$），（$\sqrt{2}$，$\sqrt{2}$）．

點(diǎn)評(píng) 本題考查矩陣與變換、逆矩陣等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力，考查化歸與轉(zhuǎn)化思想，注意解題方法的積累，屬于中檔題．