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Q.Find the area of the region lying between the parabolas y2=4axy^2 = 4ax and x2=4ayx^2 = 4ay.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 4mImportance★★★★★
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Find the intersection points, then integrate (upper curve −- lower curve) between them.

Curves: y2=4axy^2=4ax (i.e. y=2axy=2\sqrt{ax} for the upper branch) and x2=4ayx^2=4ay (i.e. y=x24ay=\dfrac{x^2}{4a}).

Points of intersection: solving simultaneously gives (0,0)(0,0) and (4a,4a)(4a,4a).

For 0≤x≤4a0\le x\le4a, y=2axy=2\sqrt{ax} (from y2=4axy^2=4ax) lies above y=x24ay=\dfrac{x^2}{4a} (from x2=4ayx^2=4ay).

Area=∫04a[2ax−x24a]dx\text{Area} = \int_0^{4a}\left[2\sqrt{ax} - \dfrac{x^2}{4a}\right]dx

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