To find area enclosed by

Y^{2} = 4ax

y = \(2\sqrt{ax}\) ....(i)

And x^{2} = 4by

y = \(\frac{x^2}{4b}\) .....(ii)

On solving the equation (i) and (ii),

\(\frac{x^4}{16b^2}\) = 4ax

Or, x^{4} – 64ab^{2}x = 0

Or, x(x^{3} – 64ab^{2}) = 0

Or, x = 0 and x = \(4\sqrt[n]{ab^2}\)

Then y = 0 and y = \(4\sqrt[n]{a^2b}\)

Equation (i) represents a parabola with vertex (0,0) and axis as x–axis,

Equation (ii) represents a parabola with vertex (0,0) and axis as x - axis,

Points of intersection of parabolas are O (0,0) and C(\(4\sqrt[n]{ab^2}\), \(4\sqrt[n]{a^2b}\))

**These are shown in the graph below: -**

The shaded region is required area, and it is sliced into rectangles of width and length (y^{1} – y^{2})ΔX.

This approximation rectangle slides from x = 0 x = 0 to x = \(4\sqrt[n]{a^2b}\) so

Required area = Region OQCPO

The area included between the parabolas y^{2} = 4ax and x^{2} = 4by is \(\frac{16}{3}\)ab sq. units