To find area enclosed by
Y2 = 4ax
y = \(2\sqrt{ax}\) ....(i)
And x2 = 4by
y = \(\frac{x^2}{4b}\) .....(ii)
On solving the equation (i) and (ii),
\(\frac{x^4}{16b^2}\) = 4ax
Or, x4 – 64ab2x = 0
Or, x(x3 – 64ab2) = 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 (y1 – y2)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 y2 = 4ax and x2 = 4by is \(\frac{16}{3}\)ab sq. units