(1)设直线l方程为y=k(x-2)
所以x^2-k^2(x-2)^2-2=0
即(k^2-1)x^2-4k^2x+4k^2+2=0
所以x1x2=(4k^2+2)/(k^2-1)
x1+x2=4k^2/(k^2-1)
y1y2=k(x1-2)*k(x2-2)
向量CA乘以向量CB=(1-x1,-y1)*(1-x2,-y2)
=x1x2-(x1+x2)+y1y2+1
=(k^2+1)x1x2-(2k^2+1)(x1+x2)+1+4k^2
=-1
(2)设M(x,y)
所以(x,y)=(x1-1,y1)+(x2-1,y2)+(-1,0)
x=x1-1+x2-1-1=4k^2/(k^2-1)-3 (1)
y=y1+y2=k[4k^2/(k^2-1)]-4k (2)
(1)式代入(2)得
x^2+2x-y^2-3=0