设A(x1,y1),B(x2,y2).
由右焦点F(√2,0),弦长为2,易求椭圆方程为x²/4+ y²/2 =1 ① ,
直线y=kx+m ②代入椭圆方程x²/4+ y²/2 =1① 得到:
(1+2k^2)x²+4kmx+2m²-4=0 ③
由题设Δ=(4km)²-8*(1+2k^2)*(m²-2) >0 且x1+x2=-4km/(1+2k^2),④y1+y2=-k(x1+x2)+2m.
线段A.B中点P在直线x+2y=0上,则(x1+x2)/2 +(y1+y2)=0,
即-2km/(1+2k^2) -4k^2m/(1+2k^2)+2m=0.解得k=1(k=-1/2舍).
|AB|=√(1+k²)*√[(x1+x2)²-4x1*x2)]=√2√[(4km/(1+2k^2)) ²- 8( m²-2)/ (1+2k^2)]=[4√(6- m²)]/(1+2k^2) =[4√(6- m²)]/3⑤
又F(√2,0)到直线y=kx+m距离x05d=|(√2+m)/√2|
∴SΔFAB=|AB|* d/2=[4√(6- m²)]/3) *|(√2+m)/√2)|/2
=√(6- m²) * (2+√2m)/ 3⑥
令[1/√(6- m²)] *(-2m)*(1+√2m/2)+√(6- m²) *√2=0
√2m²+m-3√2=0.m1=√2,m2=-3√2/2.
S(max)=(2/3)*(1+√2/√2）（√（6-2）=8/3,∴S△ABF（max)=8/3.
(S（max)=(2/3)*(1+|-3√2/2|/√2）√（6-9*2/4）=5√6/6,不是最大值)