Find the ratio of the areas of two regions of the curve `C_(1) -= 4x^(2) + pi^(2)y^(2) = 4pi^(2)` divided by the curve `C_(2) -= y = -(sgn(x - (pi)/(2)))cos x` (where sgn (x) = signum (x)).

Explanation

`4x^(2) + pi^(2)y^(2) = 4pi^(2)`
`therefore` `C_(1) : (x^(2))/(pi^(2)) + (y^(2))/(4) = 1`, which is an ellipise.
Curve `C_(2) : y = -sgn (x - (pi)/(2)) cos x`
`=({:(-cos x",", x gt (pi)/(2)),(cos x",", x lt (pi)/(2)),(0",", x = (pi)/(2)):}`
Graphs of `C_(1)` and `C_(2)` are as shown in the following figure.

From the symmetry of the curves, the areas of regions `A_(1)` and `A_(2)` are equal.
Area of ellipse is `pi(pi)(2) = 2pi^(2)`.
Area of ellipse above the x - axis is `pi^(2)`.
Now the area bounded by y = cos x and the x - axis `(A_(3))` is 2 sq. units.
So the area of the portion of the ellipse above the curve `C_(2)` is `pi^(2) - 2` (as the areas of regions `A_(1)` and `A_(2)` are same)
So the area of the portion of the ellipse below the curve `C_(2)` is `pi^(2) + 2`.
Hence the required ratio is `(pi^(2) - 2)/(pi^(2) + 2)`