a=√3,b=1,c=√2,
设直线方程为:y=kx+2,
kx-y+2=0,
设A(x1,y1),B(x2,y2)
O至直线距离d=2/√(1+k^2),
x^2/3+(kx+2)^2=1,
(1+3k^2)x^2+12kx+9=0,
根据韦达定理,
x1+x2=-12k/(1+3k^2),
x1x2=9/(1+3k^2),
根据弦长公式,
|AB|=√(1+k^2)(x1-x2)^2
=√(1+k^2)[(x1+x2)^2-4x1x2]
=√(1+k^2)[ 144k^2/(1+3k^2)^2-36/(1+3k^2)]
=[6/(1+3k^2)]√[(1+k^2)(k^2-1)],
S△OAB=d*|AB|/2=[2/√(1+k^2)]*[6/(1+3k^2)]√[(1+k^2)(k^2-1)]/2
=6√(k^2-1)/(1+3k^2),
设S=S△OAB,
t=k^2,
S=6√(t-1)/(1+3t),(1)
S^2=36(t-1)/(1+3t)^2,
设A=S^2,
9At^2+6t(A-6)+36+A=0,
当△>=0时,二次方程有实根,
36(A-6)^2-4*9A(36+A)>=0,
A