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

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(NONE) 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=

Vedclass · Accepted Answer

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

Vedclass · Answer

(NONE) 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 $m>0$,$m = \frac{2}{\sqrt{3}}$.
Note: None of the provided options match the calculated result. The correct value is $m = 2/\sqrt{3}$.

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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