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The area bounded by the parabola y2 = 4ax and x2 = 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
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)