Question

过抛物线y^2=4X焦点的弦AB , 若|AB|=8 求AB方程

Answer 1

2p=4
p/2=1
所以F(1,0)
则y=k(x-1)
所以k²(x²-2x+1)=4x
k²x²-(2k²+4)x+k²=0
x1+x2=(2k²+4)/k²
x1x2=1
则(x1-x2)²=(x1+x2)²-4x1x2=(16k²+16)/k^4
y=kx-k
所以(y1-y2)=(kx1-k-kx2+k)²=k²(x1-x2)²=(16k²+16)/k²

AB²=(x1-x2)²+(y1-y2)²=64
(16k²+16)/k^4+(16k²+16)/k²=64
k²+1+k^4+k²=4k^4
3k^4-2k²-1=0
(3k²+1)(k²-1)=0
k²=1
k=±1

所以是x+y-1=0和x-y-1=0