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Find the area included between the parabolas y2 = 4ax and x2 = 4by.

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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

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