解:(2)设A(x1,y1),B(x2,y2),y1=kx1+m,y2=kx2+m
,则弦|AB|=√(1+k^2)*︱x1-x2︱,因为k1、k、k2成等比数列,所以k^2=k1*k2,而k1=y1/x1,k2=y2/x2===>y1y2/x1x2=k^2===>
(kx1+m)(kx2+m)/(x1x2)=k^2+m[m+k(x1+x2)]/(x1x2)=k^2
===>m[m+k(x1+x2)]/(x1x2)=0
显然m<>0,否则y=kx,A,O,B三点共线,===>m/x1x2<>0===>m+k(x1+x2)=0......... (2)
将y=kx+m 代入方程x^2+4y^2-4=0,化简,得
(1+4k^2)x^2+8kmx+4(m^2-1)=0............(1)
x1,x2是方程(1)的解,===>x1+x2= -8km/(4k^2+1)代入(2)式,得m-8k^2m/(4k^2+1)=0
而m不等于0,所以 1-8k^2/(4k^2+1)=0===>4k^2=1代入(1)式并化简,得 x^2+4kmx+2(m^2-1)=0.
所以 x1+x2=-4km ,x1x2=2(m^2-1)
三角形AOB的面积S=|AB|*h/2 , h为原点O到直线AB:y=kx+m的距离,所以h=|m|/√(1+k^2),而
|AB|=√(1+k^2)*︱x1-x2︱===>S=√(1+k^2)*︱x1-x2︱*h/2=|m|*|x1-x2|/2
︱x1-x2︱=√[(x1+x2)^2- 4x1x2]=√(-4km)^2-4*2(m^2-1)=2√(2-m^2)===>S=|m|*√(2-m^2)>=√6/3
===>√(2-m^2)m^2>=√6/3===>(2-m^2)m^2>=2/3===>1-√3/3<=m^2<=1+√3/3===>
√(1-√3/3)<=m<=√(1+√3/3).
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