Question

已知點H(-3,0),點P在y軸上,點Q在x軸正半軸上,點M在直線PQ上,且·=0,又=-.?

(1)當點P在y軸上移動時,求點M的軌跡C的方程;

(2)若直線l:y=k(x-1)(k＞2)與軌跡C交于A、B兩點,AB中點N到直線3x+4y+m=0(m＞-3)的距離為,求m的取值范圍.

Answer 1

解:(1)設M(x,y).由=-,得P(0,-),Q(,0).??

由·=0(3,-)·(x,y)=0y²=4x(x＞0).                                          ?

(2)聯立y=k(x-1)(k＞2)與y²=4x,?

消去y,得k²x²-2(k²+2)x+k²=0.                                                                                  ?

由N是AB中點N(,).                                                                               ?

又由已知=|++m+3|=1.?

∵k＞2,m＞-3,∴++m+3=1.?

令=t,則0＜t＜.?

又m=-6t²-8t-2-＜m＜-2.?

結合m＞-3,∴-3＜m＜-2.