1)设椭圆方程为x^2/a^2+y^2/b^2=1 ;焦点分别为F1,F2
由题意 e=c/a=√2/2 ; PF1+PF2=2a=√6
解得a=√6/2 ,b=√3/2 ,c=√3/2 所以所求的椭圆方程为x^2/(3/4)+y^2/(3/2)=1即2x^2+4y^2=3
2)设A(x1,y1);B(x2,y2)联立x+y+m=0,2x^2+4y^2=3得6x^2+8mx+4m^2-3=0
x1+x2=-4m/3 ;x1x2=(4m^2-3)/6 ;y1y2=(-x1-m)(-x2-m)=(2m^2-3)/6
又OA垂直OB所以y1y2/x1x2=-1 即(2m^2-3)/(4m^2-3)=-1
解得m=±1