c/a=√6/3
c²/a²=2/3
2a=2√3
a=√3
a²=3
c²=2
b²=a²-c²=3-2=1
橢圓方程：x²/3+y²=1
（1）直線y=kx+1
設(shè)點(diǎn)A（x1,y1）B（x2,y2）
因?yàn)橄蛄縊A*向量OB=0
所以O(shè)A垂直O(jiān)B
y1y2+x1x2=0
(kx1+1)(kx2+1)+x1x2=0
k²x1x2+k(x1+x2)+1+x1x2=0
將y=kx+1代入x²/3+y²=1
x²+3(k²x²+2kx+1)=3
(1+3k²)x²+6kx=0
x1+x2=-6k²/(1+3k²)
x1*x2=0
-6k²/(1+3k²)+1=0
1+3k²=6k²
3k²=1
k=±√3/3
(2)直線y=kx+m原點(diǎn)到直線的距離=√3/2
那么
｜m｜/√(1+k²)=√3/2
4m²=3+3k²
將y=kx+m代入x²/3+y²=1
x²+3(k²x²+2mkx+m²)=3
(1+3k²)x²+6mkx+3m²-3=0
x1+x2=-6mk/(1+3k²)
x1*x2=(3m²-3)/(1+3k²)
AB=√(1+k²)[(x1+x2)²-4x1*x2]
因?yàn)樵c(diǎn)到直線距離為定值,所以當(dāng)AB取最大值時(shí),S就有最大值
令t=(1+k²)[36m²k²/(1+3k²)²+4(3-3m²)/(1+3k²)]
4m²=3+3k²代入
t=(1+k²)[9k²(3+3k²)/(1+3k²)²+(12-9-9k²)/(1+3k²)]
=(1+k²)(27k²+27k^4+3-9k²+9k²-27k^4)/(1+3k²)
=(1+k²)(27k²+3)/(1+3k²)²
=3(9k²+1)(k²+1)/(1+3k²)²
=3(9k^4+6k²+1+4k²)/(1+3k²)²
=3+12k²/(1+3k²)²
=3+12[k/(1+3k²)]²
=3+12[1/(1/k+3k)]²
1/k+3k≥2√3
所以t的最大值=4
AB最大值=2
S△AOB最大值=1/2×√3/2×2=√3/2