Question

Solve the following:

Find the area of the region lying between the parabolas: 4y² = 9x and 3x² = 16y

Answer 1

Given equations of the parabolas are

4y² = 9x       ...(i)

and 3x² = 16y 

∴ y = `(3x^2)/16`      ...(ii)

From (i), we get

y² = `9/4x`

∴ y = `3/2 sqrt(x)`   ...(iii)      ...[∵ In first quadrant, y > 0]

Find the points of intersection of 4y² = 9x and 3x² = 16y.

Substituting (ii) in (i), we get

`4((3x^2)/16)^2` = 9x

∴ x⁴ = 64x

∴ x⁴ – 64x = 0

∴ x(x³ – 64) = 0

∴ x = 0 or x³ = 64 = 4³

∴ x = 0 or x = 4

When x = 0, y = 0 and when x = 4, y = 3

∴ The points of intersection are O(0, 0) and B(4, 3).

Draw BD ⊥ OX.

Required area = area of the region OABCO

= area of the region ODBCO – area of the region ODBAO

= area under the parabola 4y² = 9x – area under the parabola 3x² = 16y

= `int_0^4 3/2 sqrt(x)  dx - int_0^4  (3x^2)/16  dx`

= `3/2 int_0^4 x^(1/2)  dx - 3/16  int_0^4  x^2  dx`

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

= `[(4)^(3/2) - 0] - 1/16[(4)^3 - 0]`

= `8 - 1/16 (64)`

= 8 – 4

= 4 sq.units