e=c/a=√3/2
c^2/a^2=3/4 (1)
|OM|=√5/2,AB=√5,a^2+b^2=5 (2)
a^2=b^2+c^2
a^2=4,b^2=1,c^2=3
椭圆方程x^2/4+y^2=1
设过(-1,0)的方程是y=k(x+1)
代入椭圆得
x^2/4+k^2(x+1)^2=1
(1/4+k^2)x^2+2k^2x+k^2-1=0
x1+x2=-2k^2/(1/4+k^2)
x1x2=(k^2-1)/(1/4+k^2)
PQ=√[(x1-x2)^2+(y1-y2)^2]
=√(1+k^2)*√(x1-x2)^2
=√(1+k^2)*√[(x1+x2)^2-4x1x2]
=√(1+k^2)*√[(-2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]
原点到直线的距离为d=|k|/√(1+k^2)
所以S△POQ=1/2*d*PQ
=k/2*√[(-2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]
=k/2*√[(-2k^2/(1/4+k^2))^2-4(k^2-1)/(1/4+k^2)]
=k/(1/2+2k^2)*√[4k^4-4(k^2-1)(1/4+k^2)]
=k/(1/2+2k^2)*√[4k^4-(k^2-1)(1+4k^2)]
=k/(1/2+2k^2)*√[4k^4-k^2-4k^4+1+4k^2]
=k/(1/2+2k^2)*√(3k^2+1)
=1/2*√[k^2(3k^2+1)]/(1+4k^2)^2]
令k^2=t
R=[k^2(3k^2+1)]/(1+4k^2)^2
=t(3t+1)/(1+4t)^2
(1+4t)^2*R=t(3t+1)
整理得
(16R-3)t^2+(8R-1)t+R=0
由于k^2≥0且存在
所以△=(8R-1)^2-4*(16R-3)*R≥0
64R^2-16R+1-64R^2+12R≥0
R≤1/4
因此
S=1/2*√R
≤1/2*1/2
=1/4
即面积最大值是1/4
