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Question
Find the area of the region bounded by the parabola y2 = 32x and its Latus rectum in first quadrant
Solution
Given equation of the parabola is y2 = 32x
∴ y = `+- sqrt(32x)`
∴ y = `4sqrt(2x)` ......[∵ In first quadrant, y > 0]
Required area = area of the region OBAO
= `int_0^8 y "d"x`
= `int_0^8 4sqrt(2x) "d"x`
= `4sqrt(2)[(x^(3/2))/(3/2)]_0^8`
= `(8sqrt(2))/3 [(8)^(3/2) - 0]`
= `(8sqrt(2))/3 (8sqrt(8))`
= `256/3` sq.units
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