Question

If the area enclosed between the curves $y=k{x}^2$ and $x=k{y}^2,(k>0)$, is $1$ square unit. Then $k$ is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}$

Point of intersection of the two curves are -
$x=k(k{x}^2{)}^2\Rightarrow x=0,\frac{1}{k}\Rightarrow y=0,\frac{1}{k}$

Area $=\underset{0}{\overset{\frac{1}{k}}{\int }}\frac{\sqrt{x}}{\sqrt{k}}dx-\underset{0}{\overset{\frac{1}{k}}{\int }}k{x}^{2}dx=1$
$\Rightarrow \frac{1}{\sqrt{k}}\cdot {\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{1}{k}}-k\cdot {\left[\frac{x^3}{3}\right]}_0^{\frac{1}{k}}=1$
$\Rightarrow \frac{2}{3\sqrt{k}}\cdot \frac{1}{k^{\frac{3}{2}}}-\frac{k}{3k^3}=1$
$\Rightarrow \frac{2}{3k^2}-\frac{1}{3k^2}=1$
$\Rightarrow k=\pm \frac{1}{\sqrt{3}}$
As $k>0$, so $k=\frac{1}{\sqrt{3}}$