(1)中点的轨迹方程是y=-4x,x∈(-2√5/25. 2√5/25)
(2)设交点P(x1, x1+m),Q(x2, x2+m)
且x1+x2=-2m/5. x1x2=(m^2-1)/5
OPOQ=(x1, x1+m)(x2, x2+m)=0
解得m=±√10/5
即直线l的方程是y=x±√10/5
(1)中点的轨迹方程是y=-4x,x∈(-2√5/25. 2√5/25)
(2)设交点P(x1, x1+m),Q(x2, x2+m)
且x1+x2=-2m/5. x1x2=(m^2-1)/5
OPOQ=(x1, x1+m)(x2, x2+m)=0
解得m=±√10/5
即直线l的方程是y=x±√10/5