看下边的附件 我不会这题谢谢帮忙

(I) 设M(x,y),A(x1,y1),B(x2,y2), ∵ F1(-2,0), ∴ 由已知,得
(x+2,y)=(x1+2,y1),B(x2+2,y2)+(2,0)=(x1x2+6,y1+y2)
∴ x=x1+x2+4,y=y1+y2…(*)
设AF1的方程是y=k(x+2),把它代入x²-y²=2,得
(1-k²)x²-4k²x-4k²-2=0, ∴ x1+x2=4k²/(1-k²),x1x2=-(4k²+2)/(1-k²),y1+y2=k(x1+x2)+4k=4k/(1-k²),把它代入(*)式,得x=4/(1-k²),y=4k/(1-k²),消去k,得点M的轨迹方程是x²-y²-4x=0
(II) 假设存在定点C(a,0),使向量CA•CB为常数,则
∵ y1y2=(kx1+2k)(kx2+2k)=k²x1x2+2k²(x1+x2)+4k²
=2k²/(1-k²)
(x1-a,y1)(x2-a,y2)=x1x2-a(x1+x2)+a²+y1y2
=-(4k²+2)/(1-k²)-4ak²/(1-k²)+a²+2k²/(1-k²)
=[(a²-2)-(a²+4a+2)k²]/(1-k²)
令a²+4a+2=0,则a=-2±√2,此时
(x1-a,y1)(x2-a,y2)=4(1±√2)/(1-k²)不是常数
∴ 在x轴上不存在定点C,使向量CA•CB为常数.