橢圓方程：x²/2+y²=1
a²=2,a=√2
b²=1,b=1
c²=a²-b²=2-1=1,c=1
設(shè)直線為x=my+1
斜率不存在,即直線為x=1,當(dāng)x=1時,y=±√2/2
AB=√2,S三角形AOB=1/2×1×√2=√2/2
當(dāng)斜率存在的時候
將x=my+1代入橢圓
m²y²+2my+1+2y²=2
(m²+2)y²+2my-1=0
y1+y2=-2m/(m²+2)
y1*y2=-1/(m²+2)
點(diǎn)O到直線的距離=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
當(dāng)t=1/2時,t有最大值=2,此時1/(m²+2)=1/2,m=0不合題意,因此t不能取到最大值2
也就是0