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Question
Find the area of the region lying between the parabolas 4y2 = 9x and 3x2 = 16y
Solution
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) "d"x - int_0^4 (3x^2)/16 "d"x`
= `3/2 int_0^4 x^(1/2) "d"x - 3/16 int_0^4 x^2 "d"x`
= `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
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