Question

PLEASE HELP! WILL FAN AND MEDAL
A rectangle is inscribed in the ellipse b^2x^2+a^2y^2 = a^2b^2 such that is sides are parallel of the principal axis. find the MAXIMUM perimeter.

Answer 1

Since the sides of the rectangle are parallel to the axes of the ellipse, we only need to consider the top-right vertex, whose coordinates are (x, y) with x, y ≥ 0 and x^2 / a^2 + y^2 / b^2 = 1. Then, the perimeter P = 4x + 4y. Furthermore, we can assume that b ≤ a, because if b > a, we only need to reflect everything in the line y = x.
The parametric equations of the ellipse are:
x = a*cos(t)
y = b*sin(t).
Let x_0, y_0 and t_0 be the values of x, y and t, respectively, which maximize P, i.e. P_max = 4x_0 +4y_0. Then x_0 + y_0 is also a maximum. This means that d(x+y)/dt|(x,y)=(x_0,y_0) = d(a*cos(t) + b*sin(t))/dt|t=t_0 = -a*sin(t_0) + b*cos(t_0) = 0, from which it follows that a*sin(t_0) = b*cos(t_0) and therefore sin(t_0)/cos(t_0) = tan(t_0) = b/a. Because x_0, y_0 and therefore t_0 are ≥ 0, this implies that t_0 = arctan(b/a). We thus obtain x_0 = a*cos(t_0) = a*cos(arctan(b/a)) and y_0 = b*sin(t_0) = b*sin(arctan(b/a)).

Side calculation:
Let 0 ≤ A ≤ pi/2.
Then cos(A) = 1/sec(A) = 1/sqrt(1+tan^2(A)). So, if A = arctan(r), then cos(arctan(r)) = 1/sqrt(1+tan^2(arctan(r))) = 1/sqrt(1+r^2).
Also, sin(arctan(r)) = sqrt(1-cos^2(arctan(r))) = sqrt(1-1/(1+r^2)) = sqrt(r^2/(1+r^2)) = r/sqrt(1+r^2). (end of side-calculation)

Because 0 ≤ b/a ≤ 1 < pi/2, we can apply the results of the side-calculation to our results from above:
x_0 = a*cos(arctan(b/a)) = a*1/sqrt(1+b^2/a^2) = a/sqrt(1+b^2/a^2)
y_0 = b*sin(arctan(b/a)) = b*(b/a)/sqrt(1+b^2/a^2) = b^2/(a*sqrt(1+b^2/a^2)).
P_max = 4*x_0 + 4*y_0, the required maximum perimeter.

Hope, this helps.