设抛物线y2=4x截直线y=2x+k所得弦长|AB|=3.(1)求k的值;(2)以弦AB为底边,

设抛物线y2=4x截直线y=2x+k所得弦长|AB|=3.

(1)求k的值;

(2)以弦AB为底边,x轴上的P点为顶点组成的三角形面积为39时,求点P的坐标.

答案

解:(1)设A(x1,y1)、B(x2,y2),由得4x2+4(k-1)x+k2=0,Δ=16(k-1)2-16k2>0.

∴k<.

又由韦达定理有x1+x2=1-k,x1x2=,

∴|AB|=

=·,

.∴k=-4.

(2)设x轴上点P(x,0),P到AB的距离为d,则

d=,

S△PBC=·3·=39,

∴|2x-4|=26.

∴x=15或x=-11.

∴P点为(15,0)或(-11,0).

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