设A(x1,y1),B(x2,y2)
联立L与圆的方程
y=k(x+2√2）
x²+y²=4
消去y：
(1+k²)x²+4√2k²x+8k²-4=0
Δ=(4√2k²)²-4(8k²-4)(1+k²)
=16-16k²≥0
故-1≤k≤1
由韦达定理：
x1+x2=-4√2k²/(1+k²)
x1x2=(8k²-4)/(1+k²)
|AB|²=(x1-x2)²+(y1-y2)²
=(1+k²)(x1-x2)²
=(1+k²)[(x1+x2)²-4x1x2]
=(1+k²)[32k^4/(1+k²)²-4(8k²-4)/(1+k²)]
=16(1-k²)/(1+k²)
O到L的距离d=|2√2k|/√(1+k²)
故三角形ABO面积=d*AB/2
=2|2√2k|√(1-k²)/(1+k²)
=4√2√[(1-k²)k²/(1+k²)²]
令y=[(1-k²)k²/(1+k²)²]
令k²=t
故y=[(1-t)t/(1+t)²]
=(-t²+t)/(t²+2t+1)
=-1+(3t+1)/(t²+2t+1)
=-1+3(t+1/3)/(t²+2t+1)
再利用这种方法,知道
t=1/3时,y有最大值
此时即k²=1/3
故k=±√3/3
故L:y=k(x+2√2）