Question

Find the area of the region bounded by the curves y² = 4x and 4x² + 4y² = 9 with x >= 0.

Answer 1

Required area is nothing but area bounded by the parabola y² = 4x and the circle x² + ^y2 = 9/4

To find the points of intersection.

Solving the given equations, we get

`x^2 + 4x - 9/4 = 0`

∴ 4x² + 16x − 9 = 0

∴ 4x² + 18x − 2x − 9 = 0

∴ (2x − 1)(2x + 9) = 0

∴ x = 1/2 or x  = `-9/2` (not possible)

When x = 1/2 , y = `+- sqrt2`

∴ The curves intersect at `P(1/2, sqrt2) and Q(1/2, -sqrt2)`

Consider, y² = 4x

∴ `y = 2x^(1/2) = y_1` ....(say)

Also, `x^2 + y^2 = 9/2`

`:. y^2 = 9/4 - x^2`

∴ y = `sqrt(9/4 - x^2) = y_2` ...(say)

∴ Required area = A(OPAQO) =2.A(OPAMO)

= 2[A(OPMO) + A(PAMP)]