设经过右焦点F的直线l与椭圆x^2/2+y^2=1交于A,B两点,求三角形AOB的面积最大值.O为原点

椭圆方程：x²/2+y²=1
a²=2,a=√2
b²=1,b=1
c²=a²-b²=2-1=1,c=1
设直线为x=my+1
斜率不存在,即直线为x=1,当x=1时,y=±√2/2
AB=√2,S三角形AOB=1/2×1×√2=√2/2
当斜率存在的时候
将x=my+1代入椭圆
m²y²+2my+1+2y²=2
(m²+2)y²+2my-1=0
y1+y2=-2m/(m²+2)
y1*y2=-1/(m²+2)
点O到直线的距离=1/√(1+m²)
S三角形AOB=1/2×1/√(1+m²)×√(1+m²)[(y1+y2)²-4y1y2]
=1/2*√[(4m²/(m²+2)²+4/(m²+2)]
令t=4m²/(m²+2)²+4/(m²+2)
t=(4m²+4m²+8)/(m²+2)²
=8(m²+2-1)/(m²+2)²
=8/(m²+2)-8/(m²+2)²
令u=1/(m²+2)
t=8t-8t²=-8(t²-t)=-8(t-1/2)²+2
当t=1/2时,t有最大值=2,此时1/(m²+2)=1/2,m=0不合题意,因此t不能取到最大值2
也就是0