Find the area of the region included between: y^{2} = 4ax and the line y = x

#### Solution

The vertex of the parabola y^{2} = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,

∴ y^{2} = y

∴ y^{2} – y = 0

∴ y(y – 1) = 0

∴ y = 0 or y = 1

When y = 0, x = `(0)/(2)` = 0

When y = 1, x = `(1)/(2)`

∴ the points of intersection are O(0, 0) and `"B"(1/2, 1)`

Required area = area of the region OABCO

= area of the region OABDO – area of the region OCBDO

Now, area of the region OABDO

= area under the parabola y^{2} = 4ax between x = 0 and x = `(1)/(2)`

= `int_0^(1/2)y*dx, "where" y = sqrt(2)x`

= `int_0^(1/2) sqrt(2)xdx`

= `sqrt(2)[x^(3/2)/(3/2)]_0^(1/2)`

= `sqrt(2)[2/3 (1/2)^(3/2) - 0]`

= `sqrt(2)[2/3*1/(2sqrt(2))]`

= `(1)/(3)`

Area of the region OCBDO

= area under the line y

= 4ax between x

= 0 and x = `(1)/(4ax)`

= `int_0^(1/2)y*dx, "where" y = x`

= `int_0^(1/2)2x*dx`

= `[(2x^2)/2]_0^(1/2)`

= `(4)/(3) - 0`

= `(2a^2)/(1)`

∴ required area

= `(4)/(3) = (2a^2)/(1)`

= `(8a^2)/(3) "sq units"`.