（1）c=2√2,e=c/a=√6/3
=> a=c/e=2√3,b=√(a^2-c^2)=2
∴橢圓方程為 x^2/12+y^2/4=1
（2）設(shè)直線與橢圓的交點為A(x1,y1),B(x2,y2)
設(shè)直線y=x+√2與x軸的交點為C
易求得C點的坐標(biāo)為C(-√2,0),則|CF2|=2√2-(-√2)=3√2
而S△F2AB=S△F2AC+S△F2BC
=1/2*|CF2|*|y1|+1/2*|CF2|*|y2|
=1/2*|CF2|*(|y1|+|y2)
=1/2*|CF2|*(|y1-y2|)
將直線y=x+√2代入橢圓,得
x^2/12+(x+√2)^2/4=1,整理得
2x^2+3√2x-3=0,由韋達(dá)定理有
x1+x2=-3√2/2,x1x2=-3/2; y1+y2=x1+x2+2√2=√2/2
y1y2=(x1+√2)(x2+√2)=x1x2+√2(x1+x2)+2=-3/2-3+2=-5/2
∴|y1-y2|=√(y1-y2)^2=√[(y1+y2)^2-4y1y2]
=√[(√2/2)^2-4*(-5/2)]
=√(21/2)
∴S△F2AB=1/2*|CF2|*(|y1-y2|)
=1/2*3√2*√(21/2)
=3/2*√21