已知直线y=-x加m与椭圆x2/4+y2/2=1交于A,B两点,若AB为直径的圆过原点

x²/4+y²/2=1
将y=-x+m代入得
3x²-4mx+2m²-4=0
x1+x2=4m/3
x1x2=(2m²-4)/3
AB为直径的圆过原点
即向量OA*向量OB=0
设A(x1,y1),B(x2,y2)
x1x2+y1y2=0
x1x2+(-x1+m)(-x2+m)=0
2x1x2-m(x1+x2)+m²=0
2(2m²-4)/3-4m²/3+m²=0
解得m=±2√6/3