解析几何消参
问题描述：
解析几何消参
如何由(y-2)/(t-2)=(x+2)/(t+2)
(y-2)/(t-1)=x/(t+1)
消参得到((y+1)^2)/8-((x+1)^2)/8=1 ?
答
(y-2)/(t-2)=(x+2)/(t+2)
(y-2)/(x+2)=(t-2)/(t+2)
(y-2)/(x+2)-1=(t-2)/(t+2)-1
(y-x-4)/(x+2)=-4/(t+2)
t+2=(4x+8)/(x-y+4)
t=(4x+8)/(x-y+4)-2
t=(2x+2y)/(x-y+4)
(y-2)/(t-1)=x/(t+1)
(y-2)/x=(t-1)/(t+1)
(y-2)/x-1=(t-1)/(t+1)-1
(y-x-2)/x=-2/(t+1)
t+1=2x/(x-y+2)
t=2x/(x-y+2)-1=(x+y-2)/(x-y+2)
所以(2x+2y)/(x-y+4)=(x+y-2)/(x-y+2)
(2x+2y)(x-y+2)=(x-y+4)(x+y-2)
2x²+4x-2y²+4y=x²+2x-y²+6y-8
x²+2x-y²-2y+8=0
所以(y+1)²/8-(x+1)²/8=1