导数与抛物线.已知A.B是抛物线y^2=4x上的两点,且满足OA⊥OB（O为坐标原点）,求证：（1）A,B两点的横坐标,
1个回答
(1)设A(x1,y1),B(x2,y2)
==>向量OA=(x1,y1),向量OB=(x2,y2)
OA⊥OB
==>x1x2+y1y2=0
==>y1y2=-x1x2
(y1)^2=4x1,(y2)^2=4x2
==>(y1y2)^2=16x1x2=-16y1y2
==>y1y2=-16
==>x1x2=-y1y2=16
==>A,B两点的横坐标,纵坐标之积分别为定值;
(2)(y2)^2-(y1)^2=4x2-4x1
==>(y2-y1)/(x2-x1)=4/(y1+y2)
直线AB：y-y1=(y2-y1)/(x2-x1)[x-x1]
==>y=4/(y1+y2)(x-x1)+y1
=[4(x-x1)+((y1)^2+y1y2]/(y1+y2)
=[4(x-x1)+4x1-16]/(y1+y2)
=(4x-16)/(y1+y2)
=4(x-4)/(y1+y2)
当x=4,y=0
==>直线AB经过一个定点(4,0)
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