1、设M(x,y),则(y+1)/x * (y-1)/x =-1/2,即轨迹为x^2/2+y^2=1
2、设过D(2,0)的直线为y=a(x-2),代入到轨迹C的方程中有x^2/2+a^2*(x-2)^2=1,即
(a^2+1/2)*x^2-4a^2*x+(4a^2-1)=0,有两个不同交点即方程有两个不同实根,即16a^4-4*(4a^2-1)(a^2+1/2)>0,解得 -√2/2
1、设M(x,y),则(y+1)/x * (y-1)/x =-1/2,即轨迹为x^2/2+y^2=1
2、设过D(2,0)的直线为y=a(x-2),代入到轨迹C的方程中有x^2/2+a^2*(x-2)^2=1,即
(a^2+1/2)*x^2-4a^2*x+(4a^2-1)=0,有两个不同交点即方程有两个不同实根,即16a^4-4*(4a^2-1)(a^2+1/2)>0,解得 -√2/2