The inscribed rectangle of ellipse A & # 178; / X & # 178; + B & # 178; / Y & # 178; = 1 has all fixed points on the ellipse. What is the maximum area of the rectangle?

Let point (x, y) be any point of ellipse
x=acosC
Y = bsinc (C is angle, range 0 ~ 2 π)
S (moment) = 4xy = 2absin2c
When C = 4 / π, the maximum s is 2 ab
(the equation of ellipse seems to be written wrong)

The equation of the line where one side ab of the square ABCD is located is X-Y + 4 = 0, and the vertices C and D are on the parabola y2 = x, so we can find the area of the square ABCD

Let CD equation y = x + m be substituted into y2 = X
x2+(2m-1)x+m2=0
x1+x2=1-2m
x1x2=m^2
|CD|=√2√(1-4m)
The distance between Cd and ab = | 4-m | / √ 2
So √ 2 √ (1-4m) = | 4-m | / √ 2
We get M1 = - 2, M2 = - 6
So the side length is 6 / √ 2 and the area is 18
Or the side length is 10 / √ 2 and the area is 50

In square ABCD, one edge AB is on the straight line y = x + 4, and the other two vertices C and D are on the parabola y2 = X

Let the equation of the straight line where CD is located be y = x + T, ∵ y = x + Ty2 = X. by eliminating y, X2 + (2t-1) x + T2 = 0, ∵ CD | = 2 [(1-2T) 2-4t2] = 2 (1-4t), and the distance between AB and CD is | ad | = | T-4 | 2, ∵ ad | = | CD |, | t = - 2 or - 6, so that the side length is 32 or 52. Area S1 = (32) 2 = 18, S2 = (52

The inscribed rectangle of ellipse A & # 178; / X & # 178; + B & # 178; / Y & # 178; = 1 has all fixed points on the ellipse. What is the maximum area of the rectangle?