JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Area Under Curves

The parabolas $y=k{x}^2$ and $x=k{y}^2$ meet at the origin and at $(\frac{1}{k},\frac{1}{k})$, and between them the area is ${\int }_0^{1/k}(\sqrt{\frac{x}{k}}-k{x}^2)dx$. Evaluating, the first piece gives $\frac{2}{3k^2}$ and the second $\frac{1}{3k^2}$, so the enclosed area is $\frac{2}{3k^2}-\frac{1}{3k^2}=\frac{1}{3k^2}$. Setting this to $1$ gives $k^2=\frac{1}{3}$, so $k=\frac{1}{\sqrt{3}}$.