a problem

[托斯卡尼艷陽下]

x,y為實數, 試解不等式

x^2-3xy+y^2>0

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[漫不經心]

Post by 托斯卡尼艷陽下
x,y為實數, 試解不等式
x^2-3xy+y^2>0

用旋轉來解此題， A + C = 2, A - C = -√[(1-1)^2 + (-3)^2] = -3

因此 A = -1/2, C = 5/2 => (-1/2)X^2 + (5/2)Y^2 > 0

其中 [X] = [cos(π/4) -sin(π/4)][x], 剩下只有找出雙曲線圖形的負號區
[Y] [sin(π/4) cos(π/4)][y]

和變數變換, the remaining work left to you.

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