a=√2,b=1,c=1,離心率e=c/a=√2/2,
AB的傾斜角θ=π/4,
根據(jù)焦點弦公式,
|AB|=(2b^2/a)/[1-(ecos45°)^2]
=(2*1/√2)/[1-(√2/2)^2*(√2/2)^2]
=4√2/3.
△F2AB周長=|AB|+|BF2|+|AF2|
=|AF1|+|BF1|+|BF2|+|AF2|
=(AF1|+|AF2|)+(|BF1|+|BF2|)
=2a+2a=4a=4√2.
從A作AP⊥X軸,從B作BQ⊥X軸,垂足為P、Q,
|AP|=|AF1|*cos45°,
|BQ|=|BF1|*cos45°,
|AP|+|BQ|=cos45°*|AB|=（4√2/3）*√2/2=4/3,
S△ABF2=S△AF1F2+S△BF1F2
=|F1F2|*|AP|/2+|F1F2|*|BQ|/2
=|F1F2|*|AB|/2
=2*（4/3）/2
=4/3,
∴|AB|=4/3.
△F2AB周長=4√2,
S△F2AB=4/3.