參數(shù)方程x=(t³+2t²)/(t²-1)； y=(2t³+t²)/(t²-1) 化為普通方程
y/x=(2t³+t²)/(t³+2t²)=(2t+1)/(t+2)
(t+2)y=(2t+1)x；ty-2tx=x-2y,即有(y-2x)t=x-2y,故t=(x-2y)/(y-2x).(1)；
將(1)代入y的表達(dá)式得:
y=[2(x-2y)³/(y-2x)³+(x-2y)²/(y-2x)²]/[(x-2y)²/(y-2x)²-1]
=[2(x-2y)³+(x-2y)²(y-2x)]/{[(x-2y)²-(y-2x)²](y-2x)}
=[(x-2y)²(2x-4y+y-2x)]/[(x-2y+y-2x)(x-2y-y+2x)(y-2x)]
=[-3y(x-2y)²]/[-(x+y)(3x-3y)(y-2x)]
=[-3y(x-2y)²]/[-3(x²-y²)(y-2x)]
=y(x-2y)²/(x²-y²)(y-2x)
消去y并去分母得(x-2y)²=(x²-y²)(y-2x)
故得普通方程：(x-2y)²-(x²-y²)(y-2x)=0.