以P(2,2)为圆心的圆与椭圆x^2+2y^2=m交于A,B两点,求A,B中点M的轨迹方程.
热心网友
设A(x1,y1),B(x2,y2),M(xm,ym)所以x1^2+2y1^2=m (1) x2^2+2y2^2=m (2)(2)-(1)得（x2-x1)(x2+x1)=-2(y2-y1)(y2+y1)所以kAB=(y2-y1)/(x2-x1)=-(x1+x2)/[2(y1+y2)]因为M为AB中点所以x1+x2=2xm,y1+y2=2ym所以kAB=-xm/(2ym)因为AB是圆上的弦且M为AB中点所以PM垂直AB，kPM=(ym-2)/(xm-2)所以kAB*kPM=[-xm/(2ym)]*[(ym-2)/(xm-2)]=-1所以M的轨迹方程为：xy+2x-4y=0