Question

Prove that the curves y² = 4x and x² = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

Answer 1

A(OABC) = 4 × 4 = 16 sq. units

From, y² = 4x and x² = 4y

`(x^2/4)^2 = 4x`

or     `x^4/16 = 4x`

or x⁴ - 64x = 0

or x(x³ - 64) = 0

or x = 0 or x = 4

when x = 0, y = 0

          x = 4, y = 4

Point of intersection of the two parabolas is (0, 0) and (4, 4).

Area of part III = `int_0^4 y dx` (parabola x² = 4y)

                        = `int_0^4 (x^2)/4 dx = [1/4 x^3/3]_0^4`

                        = `1/12(64 - 0)`

                        = `64/12`

                        = `16/3` sq.units

Area of I = Area of square - Area of II and III

               = `16 - int_0^4 sqrt(4x) dx`

               = `16 - (2 × 2)/3 [x^(3/2)]_0^4`

               = `16 - 32/3` sq. units

               = `16/3` sq. units

Area of II = Area of square - Area of I - Area of III

                = 16 - `16/3 - 16/3` sq. units

                = `16/3` sq. units

The two curves divide the square into three equal parts.