已知椭圆mx^2+ny^2=1与直线x+y=1相交于A,B两点,M是线段AB的中点,且/AB/=二倍根号二,OM的斜率为二分之根号二（O为坐标原点）,求m,n的值

设椭圆mx^2+ny^2=1与直线x+y-1=0交于A（x1,y1),B(x2,y2)两点A,B点在椭圆上：mx1^2+ny1^2=1mx2^2+ny2^2=1两式相减：m（x1-x2)(x1+x2)+n(y1-y2)(y1+y2)=0=> -n(y1-y2)/[m(x1-x2)]=(x1+x2)/(y1+y2)A,B也在直线上,所以：...