已知直l:y=x+m与抛物线y^2=8x交于A,B两点 ,若OA垂直于OB,求m的值.

y=x+m代入y^2=8x
(x+m)^2=8x
x^2+(2m-8)x+m^2=0
Δ=(2m-8)^2-4m^2>0 m<2
x1x2=m^2
x1+x2=8-2m
OA垂直于OB
则向量OA*向量OB=0
x1x2+y1y2=0
x1x2+(x1+m)(x2+m)=0
m^2+(x1+x2)m+2x1x2=0
m^2+(8-2m)m+2m^2=0
m^2+8m=0
m=0或-8
均满足Δ>0
但当m=0时 交点(0,0) (8,8)
O与A重合 舍去
所以y=x-8