Question

If the area between $y = mx^2$ and $x = my^2(m>0)$ 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;
$x= m\left(m x^{2}\right)^{2}=m^{3} x^{4}$
$\Rightarrow m^{3} x^{4}-x =0$
$\Rightarrow x\left(m^{3} x^{3}-1\right)=0$
$\Rightarrow x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0$
$\Rightarrow x=0, x=1 / m \text { and } y=0, y=1 / m$
We take only the points
$=(0,0) \text { and }(1 / m, 1 / m)$
Now, the area of the curve
$=\int\limits_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) d x$
Given, $\frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot x^{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m}$
$\Rightarrow \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{m^{3 / 2}}-\frac{m}{3} \cdot \frac{1}{m_{3}}\right]$
$\Rightarrow \frac{1}{4}=\left\{\frac{2}{3 m^{2}}-\frac{1}{3 m^{2}}\right\}$
$\Rightarrow \frac{1}{4}=\frac{1}{3 m^{2}}$
$\Rightarrow m^{2}=\frac{4}{3}$
$\therefore m=\pm \frac{2}{\sqrt{3}}$