Through the focus F of the ellipse x ^ 2 / 4 + y ^ 2 = 1, make the chord AB, and find the maximum area of the triangle AOB (o is the coordinate origin) Ask for detailed explanation

Through the focus F of the ellipse x ^ 2 / 4 + y ^ 2 = 1, make the chord AB, and find the maximum area of the triangle AOB (o is the coordinate origin) Ask for detailed explanation

Ellipse x ^ 2 / 4 + y ^ 2 = 1, a = 2, B = 1, C = √ 3
F1(-√3,0),F2(√3,0)
Let AB pass F1
Linear AB: y = K (x + √ 3), x = (Y - √ 3K) / K
x^2/4+y^2=1
x^2+4y^2=4
[(y-√3k)/k]^2+4y^2=4
(1+4k^2)y^2-2√3ky-k^2=0
△=(-2√3k)^2-4*(1+4k^2)*(-k^2)=16k^2*(1+k^2)
Let ya > Yb
yA-yB=√[16k^2*(1+k^2)/(1+4k^2)]
Let the area of triangle AOB (o is the coordinate origin) be s, then
S=OF1*(yA-yB)/2=0.5√3*√[16k^2*(1+k^2)]/(1+4k^2)
(16S^2-12)k^4+(8S^2-12)k^2+S^2=0
(1) Ab ⊥ X axis
x=-√3
yA-yB=1
S=√3*1/2=√3/2
(2) AB not ⊥ X axis
If the equation of unknown K ^ 2 has a real solution, then its discriminant △≥ 0, i.e
[(8S^2-12)]^2-4*(16S^2-12)*S^2≥0
S^2≤1
It can be seen that the maximum area of triangle AOB (o is the coordinate origin) is 1