If the area between y=m x^{2} and x=m y^{2} ( | vedclass

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(B) The given curves are $y=m x^{2}$ and $x=m y^{2}$ where $m>0$.To find the intersection points,substitute $y=m x^{2}$ into $x=m y^{2}$:$x=m(m x^{2})

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$\frac{2}{\sqrt{3}}$

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(B) The given curves are $y=m x^{2}$ and $x=m y^{2}$ where $m>0$.To find the intersection points,substitute $y=m x^{2}$ into $x=m y^{2}$:$x=m(m x^{2})^{2} = m^{3} x^{4}$$m^{3} x^{4}-x=0 \Rightarrow x(m^{3} x^{3}-1)=0$The intersection points are $x=0$ and $x=1/m$.The area $A$ between the curves is given by:$A = \int_{0}^{1/m} (\sqrt{x/m} - m x^{2}) dx$$A = [\frac{1}{\sqrt{m}} \cdot \frac{x^{3/2}}{3/2} - m \cdot \frac{x^{3}}{3}]_{0}^{1/m}$$A = [\frac{2}{3 \sqrt{m}} \cdot (1/m)^{3/2} - \frac{m}{3} \cdot (1/m)^{3}]$$A = \frac{2}{3 m^{2}} - \frac{1}{3 m^{2}} = \frac{1}{3 m^{2}}$Given $A = 1/4$,so $\frac{1}{3 m^{2}} = \frac{1}{4}$$3 m^{2} = 4 \Rightarrow m^{2} = 4/3$Since $m>0$,$m = \frac{2}{\sqrt{3}}$.

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

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