Question

Find the area of the region lying between the parabolas y² = 4ax and x² = 4ay.

Answer 1

The equations of the parabolas are:

`y^2=4ax`  ......(i)

`x^2=4ay`  ......(ii)

`[x^2/(4a)]^2`= 4ax  ......[by (ii)]

⇒ `x^4=64a^3x`

⇒ `x[x^3-(4a)^3]=0`

or x = 0 and x = 4a

∴ y = 0 and y = 4a

Points of intersection of curves are O(0, 0), P(4a, 4a)

∴ Required area = `int_0^(4a) [y_1 - y_2]dx`

`=int_0^(4a)sqrt(4ax)  dx-int_0^(4a)x^2/(4a)dx`

`=sqrt(4a).2/3[x^(3/2)]_0^(4a)-1/(4a) . 1/3[x^3]_0^(4a)`

`=(4sqrta)/3xx4asqrt(4a)-1/(12a)xx64a^3`

`=32/3 a^2-16/3a^2`

`=16/3a^2` sq.units