Question

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.

Answer 1

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.