Question

求直线方程

已知曲线C上任意一点M到点F(1,0)的距离比M到y轴距离大1,求c得方程.过点F的弦AB的长度为8,求直线AB的方程

Answer 1

令:M(x,y)
|MF|-|y|=1
=>(|y|+1)^2=|MF|^2
=>(1)y=1/2x^2-x(2)y=x-1/2x^2
令:y=kx-k
(1)x^2-2(k+1)x+2k=0
=>x1+x2=2k+2,x1x2=2k
=>|AB|^2=(1+k^2)(x1-x2)^2
=>k=±√3
=>y=±√3x+±√3
(2)同理：
=>y=±√3x+±√3
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