题目

已知点H(-3,0),点P在y轴上,点Q在x轴正半轴上,点M在直线PQ上,且·=0,又=-.(1)当点P在y轴上移动时,求点M的轨迹C的方程;(2)若直线l:y=k(x-1)(k＞2)与轨迹C交于A、B两点,AB中点N到直线3x+4y+m=0(m＞-3)的距离为,求m的取值范围. 答案：解:(1)设M(x,y).由=-,得P(0,-),Q(,0).由·=0(3,-)·(x,y)=0y2=4x(x＞0).                                          (2)联立y=k(x-1)(k＞2)与y2=4x,消去y,得k2x2-2(k2+2)x+k2=0.                                                                                  由N是AB中点N(,).                                                                               又由已知=|++m+3|=1.∵k＞2,m＞-3,∴++m+3=1.令=t,则0＜t＜.又m=-6t2-8t-2-＜m＜-2.结合m＞-3,∴-3＜m＜-2.

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