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

y² = 4ax

(from the other equation)

Let us solve this.

x⁴ = 64a³x

⇒ x(x³ – 64a³) = 0

⇒ x = 0 or x³ = 64a³, 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)