设M(x1,y1),N(x2,y2)
联立直线椭圆,得:
(1+2k²)x²-4k²x+2k²-4=0
x1+x2=4k²/(1+2k²),x1x2=(2k²-4)/(1+2k²)
|MN|=√[(x1-x2)²+(y1-y2)²]
=√{(x1-x2)²+[k(x1-1)-k(x2-1)]²}
=√[(x1-x2)²+k²(x1-x2)²]
=√[(1+k²)(x1-x2)²]
=√{(1+k²)[(x1+x2)²-4x1x2]
=√{(1+k²)[16k^4/(1+2k²)²-4(2k²-4)/(1+2k²)]}
=√[(1+k²)(24k²+16)/(1+2k²)²]
A点到直线距离为
h=|k|/√(1+k²)
∴S=(1/2)·h·|MN|
=(1/2)·[|k|/√(1+k²)]·√[(1+k²)(24k²+16)/(1+2k²)²]
=(1/2)·|k|·√[(24k²+16)/(1+2k²)²]
=√10/3
即:|k|·√[(24k²+16)/(1+2k²)²]=2√10/3
两边平方,得:(24k^4+16k²)/(1+2k²)²=40/9
即:7k^4-2k²-5=0
解得:k²=1或-5/7(舍去)
∴k²=1
∴k=±1