Question

椭圆C方程为(x^2)/8 +(Y^2)/4=1,若直线y=x+m与椭圆C交于不同的两点A,B,且线段AB的中点M在圆X

Answer 1

(x^2)/8 +(Y^2)/4=1

(x^2)/8 +(x+m)^2/4=1

x^2+2(x+m)^2=8

3x^2+4mx+2m^2-8=0 16m^2-24^m^2+96>0 8m^2