Question

If the area bounded by y = ax² and x = ay², a > 0, is 1 sq. units, then a =______________.

Answer 1

Given: the area bounded by y = ax² and x = ay², a > 0, is 1 sq. units

y = ax² ..(1)
x = ay² ..(2)

Solving (1) and (2), we find the coordinates of the point of intersection A.
i.e., A$\left(\frac{1}{a},\frac{1}{a}\right)$



The required area of the region OAO = Upper curve − Lower curve = 1
Thus,
$$\begin{gathered} 1={\int }_0^{\frac{1}{a}}\left(\sqrt{\frac{x}{a}}\right)dx-{\int }_0^{\frac{1}{a}}\left(a{x}^2\right)dx \\ \Rightarrow 1=\frac{1}{\sqrt{a}}{\int }_0^{\frac{1}{a}}\sqrt{x}dx-a{\int }_0^{\frac{1}{a}}{x}^{2}dx \\ \Rightarrow 1=\frac{1}{\sqrt{a}}{\left(\frac{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{3}{2}}}\right)}_0^{\frac{1}{a}}-a{\left(\frac{x^3}{3}\right)}_0^{\frac{1}{a}} \\ \Rightarrow 1=\frac{2}{3\sqrt{a}}{\left(x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_0^{\frac{1}{a}}-\frac{a}{3}{\left(x^3\right)}_0^{\frac{1}{a}} \\ \Rightarrow 1=\frac{2}{3\sqrt{a}}\left({\left(\frac{1}{a}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0\right)-\frac{a}{3}\left({\left(\frac{1}{a}\right)}^3-0\right) \\ \Rightarrow 1=\frac{2}{3a^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\left(\frac{1}{a^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)-\frac{1}{3a^2} \\ \Rightarrow 1=\frac{2}{3a^{{\displaystyle 2}}}-\frac{1}{3a^2} \\ \Rightarrow 1=\frac{2-1}{3a^{{\displaystyle 2}}} \\ \Rightarrow 1=\frac{1}{3a^{{\displaystyle 2}}} \\ \Rightarrow 3a^2=1 \\ \Rightarrow a^2=\frac{1}{3} \\ \Rightarrow a=\frac{1}{\sqrt{3}} \\ \mathrm{Thus},a=\frac{1}{\sqrt{3}} \end{gathered}$$


Hence, a = $\underline{\frac{1}{\sqrt{3}}}$.