The area bounded by the parabola y^2=4ax and x^2=4ay is:

The correct answer is Option (2) → \(\displaystyle\frac{16a^2}{3}\)

Taking a > 0, the points of intersection come from putting \(y = \displaystyle\frac{x^2}{4a}\) into \(y^2 = 4ax\):

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

    \(x(x^3 - 64a^3) = 0\)
    \(x = 0 \text{ or } x = 4a\)

So the curves meet at (0, 0) and (4a, 4a). Between them, \(y = 2\sqrt{ax}\) is the upper curve and \(y = \displaystyle\frac{x^2}{4a}\) the lower one:

    \( \displaystyle \text{Area} = \int_{0}^{4a}\left(2\sqrt{a}\,x^{1/2} - \frac{x^2}{4a}\right)dx \)

    \( \displaystyle = \left[\frac{4\sqrt{a}}{3}x^{3/2} - \frac{x^3}{12a}\right]_{0}^{4a} \)

    \( \displaystyle = \frac{4\sqrt{a}}{3}\cdot 8a\sqrt{a} - \frac{64a^3}{12a} \)

    \( \displaystyle = \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3} \)

Tip: The area between \(y^2 = 4ax\) and \(x^2 = 4by\) is \(\displaystyle\frac{16ab}{3}\), a result worth remembering; here b = a.