The area bounded between the curves $y=a x^2$ and $x=\mathrm{a} y^2(\mathrm{a}\gt0)$ is 1 sq. units, then the value of $a$ is

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Aicharya · Accepted Answer

The two curves intersect at $O(0,0)$ and $P\left(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{a}}\right)$.

According to the given condition, $\begin{aligned} & \int_0^{\frac{1}{2}}\left(\sqrt{\frac{x}{\mathrm{a}}}-\mathrm{a} x^2\right) \mathrm{d} x=1 \\ & \Rightarrow\left[\frac{2}{3 \sqrt{\mathrm{a}}} x^{3 / 2}-\frac{\mathrm{a} x^3}{3}\right]_0^{1 / \mathrm{a}}=1 \\ & \Rightarrow \frac{2}{3 \sqrt{\mathrm{a}}} \times \frac{1}{\mathrm{a}^{3 / 2}}-\frac{\mathrm{a}}{3} \times \frac{1}{\mathrm{a}^3}=1 \\ & \Rightarrow \frac{2}{3 \mathrm{a}^2}-\frac{1}{3 \mathrm{a}^2}=1 \Rightarrow \frac{1}{3 \mathrm{a}^2}=1 \\ & \Rightarrow \mathrm{a}=\frac{1}{\sqrt{3}} \quad \ldots[\because \mathrm{a}\gt0] \end{aligned}$

The area bounded between the curves $y=a x^2$ and $x=\mathrm{a} y^2(\mathrm{a}\gt0)$ is 1 sq. units, then…

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