后续
1
(1)由 y=2x+b
y2=4x
,消去y得4x2+4(b-1)x+b2=0.
△=[4(b-1)]2-4×4×b2>0,得b<1
2
.
x1+x2=1-b,x1•x2=b2
4
.
|AB|= (1+22)[(x1+x2)2-4x1x2]
= 5
(1-b)2-b2
=3 5
.
∴解得:b=-4,满足b<1
2
,∴b=-4;
(2)P到直线2x-y-4=0的距离为d,d=|2x0-4|
5
.
由S△PAB=1
2
×3 5
×|2x0-4|
5
=9,解得:x=5或x=-1,
∴P点坐标为(-1,0)或(5,0).