题目内容

如图,P(-3,0),点A在y轴上,点Q在x轴的正半轴上,且·=0,在AQ的延长线上取一点M,使||=2||.

(1)当A点在y轴上移动时,求动点M的轨迹C的方程;

(2)已知k∈R,i=(0,1),j=(1,0),经过(-1,0)以ki+j为方向向量的直线l与轨迹C交于E、F两点,又点D(1,0),若∠EDF为钝角时,求k的取值范围.

解:(1)设A(0,y0)、Q(x0,0)、M(x,y),

=(-3,-y0),=(x0,-y0).

·=0,∴-3x0+(-y0)(-y0)=0.

∴y02=3x0.                                                                ① 

又||=2||,

                                             ②

将②代入①,有y2=4x(x≠0).                                                      

(2)ki+j=k(0,1)+(1,0)=(1,k),

则l:y=k(x+1),与y2=4x联立,

得k2x2+(2k2-4)x+k2=0,

x1+x2=,x1x2=1.

当Δ>0时,k∈(-1,0)∪(0,1),                                                 ③ 

=(x1-1,y1),=(x2-1,y2),

若∠EDF为钝角,则·<0.                                               

·=(x1-1)(x2-1)+y1y2=x1x2-(x1+x2)+k(x1+1)k(x2+1)+1

=(k2+1)x1x2+(k2-1)(x1+x2)+k2+1<0,                                          ④ 

将③代入④整理有4k2-2<0.

<k<.

由题知k≠0,

∴满足题意k∈(,0)∪(0,).

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