Question

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

• A. $\pm 3\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\pm \frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

Given curves; $y=m{x}^2$ and $y^{2}m=x;m>0$
Intersection point of both curves;
$x=m{\left(m{x}^2\right)}^2=m^3{x}^4$
$\Rightarrow m^3{x}^4-x=0$
$\Rightarrow x\left(m^3{x}^3-1\right)=0$
$\Rightarrow x(mx-1)\left(m^2{x}^2+1+mx\right)=0$
$\Rightarrow x=0,x=1/m\text{ and }y=0,y=1/m$
We take only the points
$=(0,0)\text{ and }(1/m,1/m)$
Now, the area of the curve
$=\underset{0}{\overset{1/m}{\int }}\left(\sqrt{\frac{x}{m}}-m{x}^2\right)dx$
Given, $\frac{1}{4}={\left[\frac{2}{3\sqrt{m}}\cdot x^{3/2}-m\cdot \frac{x^3}{3}\right]}_0^{1/m}$
$\Rightarrow \frac{1}{4}=\left[\frac{2}{3\sqrt{m}}\cdot \frac{1}{m^{3/2}}-\frac{m}{3}\cdot \frac{1}{m_3}\right]$
$\Rightarrow \frac{1}{4}=\left\{\frac{2}{3m^2}-\frac{1}{3m^2}\right\}$
$\Rightarrow \frac{1}{4}=\frac{1}{3m^2}$
$\Rightarrow m^2=\frac{4}{3}$
$\therefore m=\pm \frac{2}{\sqrt{3}}$