The area bounded by the parabola y^{2} = 4ax and x^{2} = 4ay is

This problem is the generalized form of question 2.

Let’s proceed to solve it similarly –

We need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.

This means that at points of intersection

y^{2} = 4ax

(from the other equation)

Let us solve this.

x^{4} = 64a^{3}x

⇒ x(x^{3} – 64a^{3}) = 0

⇒ x = 0 or x^{3} = 64a^{3}, i.e, x = 4a

So the points of intersection are (0,0) and (4a,4a)

Now, let’s compute the area.

If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4a.

i.e.,

(Ans)

1