Question

Solve the following :

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

Answer 1

For finding the points of intersection of the two parabolas, we equate the values of 4y² from their equations.
From the equation 3x² = 16y, y = `(3x^2)/(16)`

∴ y = `(3x^4)/(256)`

∴ `(3x^4)/(256)` = 9x

∴ 3x⁴ – 2304x = 0
∴ x(x³ – 2304) = 0
∴ x = 0 or x³ = 2304

i.e. x = 0 or x = 4

When x = 0, y = 0

When x = 4, y = `(4^2)/(4)` = 4

∴ the points of intersection are O(0, 0) and A(4, 4).

Required area = area of the region OBACO

= [area of the region ODACO] –  [area of the region ODABO]

Now, area of the region ODACO

= area under the parabola y² = 4x,

i.e. y = `2sqrt(x)` between x = 0 and x = 4

= `int_0^4 2sqrt(x)*dx`

= `[2  (x^(3/2))/(3/2)]_0^4`

= `2 xx (2)/(3) xx 4^(3/2) - 0`

= `(4)/(3) xx (2^3)`

= `(32)/(3)`
Area of the region ODABO

= area under the parabola x² = 4y,

i.e. y = `x^2/(4)` between x = 0 and x = 4

= `int_0^4 (1)/(4)x^2*dx`

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

= `(1)/(4)(64/3 - 0)`

= `(16)/(3)`

∴ required area = `(32)/(3) - (16)/(3)`

= 4 sq units"`.