設橢圓方程為x²/a²+y²/b²=1
c/a=√2/2
c²/a²=1/2
a²=2c²
a²=b²+c²
所以b²=c²
橢圓方程：x²/2b²+y²/b²=1
x²+2y²=2b²
設直線為y=kx+2
代入
x²+2(k²x²+4kx+4)=2b²
(1+2k²)x²+8kx+8-2b²=0
x1+x2=-8k/(1+2k²)
x1*x2=(8-2b²)/(1+2k²)
原點到直線AB的距離d=2/√(1+k²)
S△AOB=1/2×AB×d=1/√(1+k²)×√(1+k²)[(x1+x2)²-4x1x2]
令t=(x1+x2)²-4x1x2=64k²/(1+2k²)²+4(2b²-8)/(1+2k²)
=(64k²+8b²-32+16k²b²-64k²)/(1+2k²)²
=[8b²(1+2k²)-32]/(1+2k²)²
=8b²/(1+2k²)-32/(1+2k²)²
令u=1/(1+2k²)
t=8b²u-32u²
=-32(u²-b²u/4)
=-32(u-b²/8)²+b^4/2
t為二次函數(shù)
當u=b²/8時,t有最大值b^4/2
S=b²/√2
根據(jù)題意
b²/√2=√2
b²=2
a²=2b²=4
橢圓方程：x²/4+y²/2=1
此時1/(1+2k²)=2/8
1+2k²=4
k²=3/2
k=±√6/2
直線方程：y=±√6/2x+2
即±√6x-2y+4=0