若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2则M/N的值等于

椭圆mx^2+ny^2=1与直线x+y-1=0交于A,B两点,过原点与线段AB中点的直线斜率为√2/2,则n/m=____√2______
联立椭圆与直线方程得到：
mx^2+ny^2-1=0
y=1-x
===> mx^2+n(x-1)^2-1=0
===> mx^2+nx^2-2nx+n-1=0
===> (m+n)x^2-2nx+(n-1)=0
===> x1+x2=2n/(m+n)
而，y1+y2=(1-x1)+(1-x2)=2-(x1+x2)=2-[2n/(m+n)]=2m/(m+2n)
所以，AB中点C的坐标为：
Cx=(x1+x2)/2=n/(m+n)
Cy=(y1+y2)/2=m/(m+n)
所以，OC所在直线的斜率Koc=(Cy-0)/(Cx-0)=Cy/Cx
=[m/(m+n)]/[n/(m+n)]
=m/n
=√2/2
所以，n/m=1/(√2/2)=√2

跟2/2

点差法：设A(x1,y1),B(x2,y2),AB中点C(x0,y0),则2x0=x1+x2,2y0=y1+y2；OC的斜率k=y0/x0=(y1+y2)/(x1+x2)=√2/2因为A,B在椭圆上,所以,mx1²+ny1²=1mx2²+ny2²=1点差得：m(x1²-x2²)+n(y1&...