题目内容
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.
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(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.
P点为(15,0)或(-11,0).
(1)设A(x1,y1)、B(x2,y2),
由
得4x2+4(k-1)x+k2=0,Δ=16(k-1)2-16k2>0.
∴
.
又由韦达定理有x1+x2=1-k,x1x2=
,
∴|AB|=
=
,
即
.
∴k=-4.
(2)设x轴上点P(x,0),P到AB的距离为d,则
,
S△PBC=
·
·
=39,
∴|2x-4|=26.
∴x=15或x=-11.
∴P点为(15,0)或(-11,0).
由
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∴
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又由韦达定理有x1+x2=1-k,x1x2=
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∴|AB|=
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=
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即
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∴k=-4.
(2)设x轴上点P(x,0),P到AB的距离为d,则
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S△PBC=
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∴|2x-4|=26.
∴x=15或x=-11.
∴P点为(15,0)或(-11,0).
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