问题描述: 若椭圆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,求n/m的值 1个回答 分类:数学 2014-12-01 问题解答: 我来补答 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:(y1-y2)/(x1-x2)=直线斜率=-1=> n/m=(x1+x2)/(y1+y2)令A,B的中点为(x0,y0)=> x0=(x1+x2)/2 ; y0=(y1+y2)/2=> n/m=x0/y0 = (x0-0)/(y0-0)中点到原点直线的斜率的倒数=> n/m = √2 展开全文阅读