設圓C：X^2+Y^2=4經矩陣A線性變換後所得到的新圖形

R：X^2+Y^2=4 經矩陣 A＝(1 -2; 0 1) 線性變換後得到新圖形T (1) 求T與原點的最大距離?
T = A*R ==> R = A^(-1)*TA^(-1) = 反矩陣 = [1 2; 0 1]{X}.[1 2]{x}.{x+2y}
{Y}=[0 1]{y}={.y..}4 = X^2 + Y^2= (x+2y)^2 + y^2 = x^2 + 4xy + 5y^2
D = 判別式=|1 2|
.|2 5|= 5 - 4= 1 > 0= 橢圓形
作第2次線性轉換,把橢圓旋轉到正位:Q = 旋轉角度= 0.5*atan[4/(1-5)]= 0.5*atan(-1)= 0.5*(-45) deg= -22.5 degc = cos(-22.5) = √[(2+√2)/2]s = sin(-22.5) = -√[(2-√2)/2]sin(-45) = -√2/2
A' = x'^2係數= A*c^2 + B*sin(2Q) + C*s^2= 1*[(2+√2)/2] - 4*√2/2 + 5*[(2-√2)/2]= 6 - 4√2
C' = y'^2係數= A*s^2 - B*sin(2Q) + C*c^2= 1*[(2-√2)/2] + 4*√2/2 + 5*[(2+√2)/2]= 6 + 4√2
(6-4√2)x'^2 + (6+4√2)y'^2 = 4a = 半長軸 ;;; y'=0= 2/√(6 - 4√2)= √(6 + 4√2)
T與原點的最大距離 = √(6 + 4√2)
(2) (-4,2)與(4,-2)是否為T的其中兩個頂點？ F(x,y) = x^2 + 4xy + 5y^2 - 4F(-4,2) = F(4,-2)= 16 - 4*4*2 + 20 - 4= 36 - 36= 0= 兩點在橢圓形上
Q = atan(-2/4)= atan(-1/2)=\= 22.5 deg= 否定

2015-05-27 10:15:23 補充：
修正B的數據: 2B = 4 ==> B = 2

A' = A*c^2 + B*sin(2Q) + C*s^2

= (2+√2)/2 - 2*√2/2 + 5*[(2-√2)/2]

= 6 - 3√2


C' = A*s^2 - B*sin(2Q) + C*c^2

= (2-√2)/2 + 4*√2/2 + 5*[(2+√2)/2]

= 6 + 3√2


a = 2/√(6 - 3√2)

= √[6 (2 + √2)] / 3

= T與原點的最大距離

2015-05-27 10:16:41 補充：
4*√2/2 改為 2*√2/2