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(A) The two curves $y=ax^2$ and $x=ay^2$ intersect at $O(0,0)$ and $P\left(\frac{1}{a}, \frac{1}{a}\right)$.To find the area bounded by these curves,w

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$\frac{1}{\sqrt{3}}$

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(A) The two curves $y=ax^2$ and $x=ay^2$ intersect at $O(0,0)$ and $P\left(\frac{1}{a}, \frac{1}{a}\right)$.To find the area bounded by these curves,we integrate the difference between the upper curve $y=\sqrt{\frac{x}{a}}$ and the lower curve $y=ax^2$ from $x=0$ to $x=\frac{1}{a}$.Given area $= \int_0^{\frac{1}{a}} \left(\sqrt{\frac{x}{a}} - ax^2\right) dx = 1$$\Rightarrow \left[\frac{1}{\sqrt{a}} \cdot \frac{x^{3/2}}{3/2} - \frac{ax^3}{3}\right]_0^{\frac{1}{a}} = 1$$\Rightarrow \left[\frac{2}{3\sqrt{a}} x^{3/2} - \frac{ax^3}{3}\right]_0^{\frac{1}{a}} = 1$Substituting the limits:$\Rightarrow \left(\frac{2}{3\sqrt{a}} \cdot \left(\frac{1}{a}\right)^{3/2} - \frac{a}{3} \cdot \left(\frac{1}{a}\right)^3\right) = 1$$\Rightarrow \frac{2}{3\sqrt{a} \cdot a\sqrt{a}} - \frac{a}{3a^3} = 1$$\Rightarrow \frac{2}{3a^2} - \frac{1}{3a^2} = 1$$\Rightarrow \frac{1}{3a^2} = 1$$\Rightarrow a^2 = \frac{1}{3}$Since $a > 0$,we have $a = \frac{1}{\sqrt{3}}$.

The area bounded between the curves $y=ax^2$ and $x=ay^2$ $(a > 0)$ is $1$ sq. unit. Then the value of $a$ is:

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