Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option.

Step 1: Find the point of intersection of the curves

Given curves are, $y=m{x}^2$ and $x=m{y}^2(m>0)$.

The point of intersection of both curves:
$\begin{array}{rcl}\therefore x& =& m{\left(m{x}^2\right)}^2\\ & \Rightarrow & x=m^3{x}^4\\ & \Rightarrow & m^3{x}^4-x=0\\ & \Rightarrow & x\left(m^3{x}^3-1\right)=0\end{array}$

Using ${\mathit{a}}^{\mathbf{3}}\mathbf{-}{\mathit{b}}^{\mathbf{3}}\mathbf{=}\left(a-b\right)\left(a^2+ab+b^2\right)$ we have:

$\begin{array}{rcl}x(mx-1)\left(m^2{x}^2+1+mx\right)& =& 0\end{array}$

Now using the zero property of product we get:

$\begin{array}{rcl}x& =& 0,x=\frac{1}{m}\text{and}y=0,y=\frac{1}{m}\end{array}$

So, the point of intersection are: $\left(0,0\right),\left(\frac{1}{m},\frac{1}{m}\right)$.

Step 2: Find the area

The area under the curve is given by:

$\begin{array}{rcl}A& =& {\int }_0^{1/m}\left(\sqrt{\frac{x}{m}}-m{x}^2\right)dx\\ & =& {\left[\frac{2}{3\sqrt{m}}·x^{3/2}-m·\frac{x^3}{3}\right]}_0^{1/m}\\ & =& \left[\frac{2}{3\sqrt{m}}·\frac{1}{m^{3/2}}-\frac{m}{3}·\frac{1}{m^{3^{}}}\right]\\ & =& \left\{\frac{2}{3m^2}-\frac{1}{3m^2}\right\}\\ & =& \frac{1}{3m^2}\end{array}$

As the area between the curves is $\frac{1}{4}\mathrm{sq}\text{units}$,

$\begin{array}{rcl}\therefore \frac{1}{4}& =& \frac{1}{3m^2}\\ & \Rightarrow & m^2=\frac{4}{3}\\ & \Rightarrow & m=\pm \frac{2}{\sqrt{3}}\end{array}$

Hence, option B is correct.