(1)由OA丄OB,可得x1x2+y1y2=0 ∵y12=4x1,y22=4x2,∴16x1x2=(y1y2)2 代入上式得16y1y2+(y1y2)2=0 ∵y1y2≠0,∴y1y2=-16,∴x1x2=16; (2)设T(t,0),当x1≠x2时,A,B,T三点共线,∴= ∴(y2-y1)t=y2x1-y1x2=-4(y1-y2) ∵y1≠y2,∴t=4 当x1=x2时,∵OA⊥OB,此时△AOB为等腰直角三角形,x1=x2=t,直线OA的方程式为y=x 与抛物线联立,解得t=x1=4 ∴T的坐标是(4,0); (3)设R(x,y),由F(1,0),+=,得(x1-1,y1)+(x2-1,y2)=(x-1,y) 即 ∵y12=4x1,y22=4x2,∴两式相减可得(y1-y2)(y1+y2)=4(x1-x2) 当x1≠x2时,y•=4 ∵AB的中点M(,),点T(4,0)都在直线AB上, ∴kAB=kTM,即=代入上式得y•=4 化简可得y2=4x-28 当x1=x2时,点R(7,0)符合上式 综上可知点R的轨迹方程是y2=4x-28. |