（1）kx-y+b=0
b/√(1+k2)=1
f(k)=√(1+k2)
(2) A(x1,y1) B(x2,y2)
x1x2+y1y2=2/3
把y=kx+b代入x^2/2+y^2=1
(1+2k2)x2+4bkx+2b2-2=0
x1x2=(2b2-2)/(1+2k2)=2k2/(1+2k2)
x1+x2=-4bk/(1+2k2)
y1y2=k^2x1x2+kb(x1+x2)+b2
=2k^4/(1+2k2)-4k2b2/(1+2k2)+b2
所以
2k^4/(1+2k2)-4k2b2/(1+2k2)+b2+2k2/(1+2k2)=2/3
解得k=±1 ,b=√2
所以直線l的方程y=±x+√2
(3)
2k^4/(1+2k2)-4k2b2/(1+2k2)+b2+2k2/(1+2k2)=m
(k2+1)/(2k2+1)=m
2/3≤m≤3/4
2/3≤(k2+1)/(2k2+1)≤3/4
得1/2≤k2≤1
又由(1+2k2)x2+4bkx+2b2-2=0
x1x2=(2b2-2)/(1+2k2)=2k2/(1+2k2)
x1+x2=-4bk/(1+2k2)
得|AB|=2√2√(1+k2)√k2/(1+2k2)
高為1
所以面積S=1/2*2√2√(1+k2)√k2/(1+2k2)
因?yàn)?/2≤k2≤1
所以S的范圍為