Question

The area bounded by the parabola y² = 4ax and x² = 4ay is
(a) $\frac{8a^3}{3}$

(b) $\frac{16a^2}{3}$

(c) $\frac{32a^2}{3}$

(d) $\frac{64a^2}{3}$

Answer 1

$\left(\mathrm{b}\right)\frac{16a^2}{3}$


To find the point of intersection of the parabolas substitute $y=\frac{x^2}{4a}$ in $y^2=4ax$ we get
$$\begin{gathered} \frac{x^4}{16a^2}=4ax \\ \Rightarrow x^4-64a^{3}x=0 \\ \Rightarrow x\left(x^3-64a^3\right)=0 \\ \Rightarrow x=0\mathrm{or}x=4a \\ \Rightarrow y=0\mathrm{or}y=4a \end{gathered}$$
Therefore, the required area ABCD,
$$\begin{gathered} A={\int }_0^{4a}\left(y_1-y_2\right)\mathit{d}x\left(\mathrm{Where},y_1=2\sqrt{ax}\mathrm{and}{y}_2=\frac{x^2}{4a}\right) \\ ={\int }_0^{4a}\left(2\sqrt{ax}-\frac{x^2}{4a}\right)\mathit{d}x \\ ={\left[\frac{4\sqrt{a}}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{x^3}{12a}\right]}_0^{4a} \\ =\left[\frac{4\sqrt{a}}{3}{\left(4a\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{{\left(4a\right)}^3}{12a}\right]-\left[\frac{4\sqrt{a}}{3}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{{\left(0\right)}^3}{12a}\right] \\ =\left[\frac{4\sqrt{a}}{3}8a^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{64a^3}{12a}\right]-0 \\ =\frac{32a^2}{3}-\frac{16a^2}{3} \\ =\frac{16a^2}{3}\mathrm{square}\mathrm{units} \end{gathered}$$