If the area between \(y=m x^{2}\) and \(x=m y^{2}(m>0)\) is \(1 / 4 \mathrm{sq}\) units, then the value of \(m\) is

Given curves; \(y=m x^{2}\) and \(y^{2} m=x ; m>0\)
Intersection point of both curves;
\(x=m\left(m x^{2}\right)^{2}=m^{3} x^{4}\)
\(\Rightarrow \quad m^{3} x^{4}-x=0\)
\[
\begin{array}{cc}
\Rightarrow & x\left(m^{3} x^{3}-1\right)=0 \\
\Rightarrow & x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0 \\
\Rightarrow & x=0, x=1 / m \text { and } y=0, y=1 / m
\end{array}
\]
We take only the points
\[
=(0,0) \text { and }(1 / m, 1 / m)
\]
Now, the area of the curve
\[
\begin{aligned}
&=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) d x \\
\text { Given, } \quad \frac{1}{4} &=\left[\frac{2}{3 \sqrt{m}} \cdot x^{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \\
\Rightarrow \quad \frac{1}{4} &=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{m^{3 / 2}}-\frac{m}{3} \cdot \frac{1}{m_{3}}\right] \\
\Rightarrow \quad \frac{1}{4} &=\left\{\frac{2}{3 m^{2}}-\frac{1}{3 m^{2}}\right\} \\
\Rightarrow \quad \frac{1}{4} &=\frac{1}{3 m^{2}} \Rightarrow m^{2}=\frac{4}{3} \\
\therefore \quad m=\pm \frac{2}{\sqrt{3}}
\end{aligned}
\]