#### Question

Find the area of the region bounded by the curves y^{2} = 4x and 4x^{2} + 4y^{2} = 9 with x >= 0.

#### Solution

Required area is nothing but area bounded by the parabola y^{2} = 4x and the circle x^{2} + ^{y2} = 9/4

To find the points of intersection.

Solving the given equations, we get

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

∴ 4x^{2} + 16x − 9 = 0

∴ 4x^{2} + 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, y^{2} = 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)]

Is there an error in this question or solution?

#### APPEARS IN

Solution Find the Area of the Region Bounded by the Curves Y2 = 4x and 4x2 + 4y2 = 9 with X >= 0. Concept: Pair of Straight Lines - Point of Intersection of Two Lines.