Question

Find the area of the region bounded by the parabola y² = 32x and its Latus rectum in first quadrant

Answer 1

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