解:设P(2,-1)的直线为y=k(x-2)-1代入双曲线2xˇ2-yˇ2=2得,(2-k2)x2+(4k2+2k)x-(4k2+4k+3)=0,2-k2≠0,△>0恒成立,设A(x1,y1),B(x2,y2),M(x,y)于是x1+x2=(4k2+2k)/(k2-2),y1+y2=(8k+4)/(k2-2)。即2x=(4k2+2k)/(k2-2)....(#),2y=(8k+4)/(k2-2).....(##) ,两式相除得,k=2x/y,代入(##)得2x2-4x-y2-y=0,当直线与X轴垂直时...
|