JEE Main 2019 — Area Under Curves Question with Solution

The given curves are, $y=k{x}^2, x=k{y}^2$

To find the point of intersection of the curves, put $y$ from the first curve into second, to get

$x=k\left(k^2{x}^4\right)$

$\Rightarrow x=0$ or $x^3={\left(\frac{1}{k}\right)}^3\Rightarrow x=\frac{1}{k}$

Hence, $y=0$ or $y=k{\left(\frac{1}{k}\right)}^2=\frac{1}{k}$

Therefore, point of intersection is $\left(\frac{1}{k}, \frac{1}{k}\right).$

Hence, the required area is ${\int }_0^{\frac{1}{k}}\left(y_2-y_1\right)dx=1$

$\Rightarrow {\int }_0^{\frac{1}{k}}\left(\sqrt{\frac{x}{k}}-k{x}^2\right)dx=1$

$\Rightarrow {\left(\frac{1}{\sqrt{k}}\frac{x^{3/2}}{3/2}-\frac{k{x}^3}{3}\right)}_0^{\frac{1}{k}}=1$

$\Rightarrow \frac{2}{3k^2}-\frac{1}{3k^2}=1$

$\Rightarrow k^2=\frac{1}{3}$

$\Rightarrow k=\pm \sqrt{\frac{1}{3}}$

But, given $k>0$

So, $k=\frac{1}{\sqrt{3}}.$