过点(2,-1)作直线交双曲线2X^2-Y^2=2于P、Q两点,求线段PQ的中点M的轨迹方程

直线斜率为k,P(x1,y1),Q(x2,y2),M(m,n)2x1^2-y1^2=2 1式2x2^2-y2^2=2 2式1式减2式,得2(x1^2-x2^2)=y1^2-y2^22[(x1+x2)/(y1+y2)]=(y1-y2)/(x1-x2)2m/n=k又k=(n+1)/(m-2)就可得出结果了