若椭圆ax^2+by^2=1（a>0,b>0)与直线x+y+1=0交于AB两点,M为AB的中点,直线OM的斜率为2,且OA垂直于OB

xA+xB=2xM,yA+yB=2yM
k(AB)=(yA-yB)/(xA-xB)=-1,b=1,k(OM)=yM/xM=(yA+yB)/(xA+xB)=2
ax^2+by^2=1
[a(xA)^2+b(A)^2]-[a(xB)^2+b(yB)^2]=0
a(xA+xB)*(xA-xB)+b(yA+yB)/(xA-xB)=0
a+b[(yA+yB)/(xA+xB)]*[(yA-yB)/(xA-xB)]=0
a+b*2*(-1)=0
a=2b
OA垂直于O（O为坐标原点）?
OA肯定不垂直于OB（O为坐标原点）因为k(OM)*k(AB)不等于－1
条件不足