Question

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

Answer 1

To find the point of intersection of the parabolas substitute $y=\frac{x^2}{4b}$ in $y^2=4ax$ we get

$$\begin{gathered} \frac{x^4}{16b^2}=4ax \\ \Rightarrow x^4-64a{b}^{2}x=0 \\ \Rightarrow x\left(x^3-64a{b}^2\right)=0 \\ \Rightarrow x=0\mathrm{and}x=4\sqrt[3]{a{b}^2} \\ \Rightarrow y=0\mathrm{and}y=4\sqrt[3]{a^{2}b} \end{gathered}$$

Therefore, the required area ABCD = ${\int }_0^{4\sqrt[3]{a{b}^2}}\left(y_1-y_2\right)\mathit{d}x$ where $y_1=2\sqrt{ax}$ and $y_2=\frac{x^2}{4b}$.

Required area =
$$\begin{gathered} {\int }_0^{4\sqrt[3]{a{b}^2}}\left(y_1-y_2\right)\mathit{d}x \\ ={\int }_0^{4\sqrt[3]{a{b}^2}}\left(2\sqrt{ax}-\frac{x^2}{4b}\right)\mathit{d}x \\ ={\left[\frac{4\sqrt{a}}{3}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{x^3}{12b}\right]}_0^{4\sqrt[3]{a{b}^2}} \\ =\left[\frac{4\sqrt{a}}{3}{\left(4\sqrt[3]{a{b}^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{{\left(4\sqrt[3]{a{b}^2}\right)}^3}{12b}\right]-\left[\frac{4\sqrt{a}}{3}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{{\left(0\right)}^3}{12a}\right] \\ =\frac{16ab}{3}\mathrm{square}\mathrm{units} \end{gathered}$$