Question

如圖,P(-3,0),點A在y軸上,點Q在x軸的正半軸上,且·=0,在AQ的延長線上取一點M,使||=2||.

(1)當A點在y軸上移動時,求動點M的軌跡C的方程;

(2)已知k∈R,i=(0,1),j=(1,0),經過(-1,0)以ki+j為方向向量的直線l與軌跡C交于E、F兩點,又點D(1,0),若∠EDF為鈍角時,求k的取值范圍.

Answer 1

解：(1)設A(0,y₀)、Q(x₀,0)、M(x,y),

則=(-3,-y₀),=(x₀,-y₀).

又·=0,∴-3x₀+(-y₀)(-y₀)=0.

∴y₀²=3x₀.                                                                ① 

又||=2||,

∴∴                                             ②

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

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

則l:y=k(x+1),與y²=4x聯立,

得k²x²+(2k²-4)x+k²=0,

x₁+x₂=,x₁x₂=1.

當Δ＞0時,k∈(-1,0)∪(0,1),                                                 ③ 

又=(x₁-1,y₁),=(x₂-1,y₂),

若∠EDF為鈍角,則·＜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 

將③代入④整理有4k²-2＜0.

∴＜k＜.

由題知k≠0,

∴滿足題意k∈(,0)∪(0,).