Question

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.

Answer 1

解：(1)设A(0,y₀)、Q(x₀,0)、M(x,y),

则=(-3,-y₀),=(x₀,-y₀).

又·=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 

又=(x₁-1,y₁),=(x₂-1,y₂),

若∠EDF为钝角,则·＜0.                                               

而·=(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,).