Question

椭圆x^2/2^2+y^2=1与直线x-y+m=0交于不同两点A,B 且直线AB中点不在圆x^2+y^2=5/9内,求m

Answer 1

x^2/2^2+y^2=1

x^2+4y^2=4

x-y+m=0

x=y-m

代入椭圆方程得

y^2-2my+m^2+4y^2-4=0

5y^2+-2my+m^2-4=0

(y1-y2)^2=(y1+y2)^2-4y1y2=(2m/5)^2-4/5(m^2-4)=-16/25m^2+16/5

AB中点y^2=1/4(y1-y2)^2=-4/25m^2+4/5=x^2

故x^2+y^2==-2/25m^2+2/50

两个不等式联立,解得m即可