Question

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

Answer 1

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=1
mx2^2+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