The area of the region bounded by the curves x (1+ y 2 )=1 and y 2 =2 x is :

Mathematics · Area under the Curves

JEE Main 2025 — 28 January, Evening Shift — Question 9

The area of the region bounded by the curves $\mathrm{x}\left(1+\mathrm{y}^{2}\right)=1$ and $\mathrm{y}^{2}=2 \mathrm{x}$ is :

$\mathrm{x}\left(1+\mathrm{y}^{2}\right)=1..(1)$

$y^{2}=2 x…(2)$

From equation (1) & (2)

$\mathrm{x}(1+2 \mathrm{x})=1 \Rightarrow 2 \mathrm{x}^{2}+\mathrm{x}-1=0$

$\Rightarrow \mathrm{x}=\frac{1}{2}, \mathrm{x}=-1$ (Reject)

$\Rightarrow \mathrm{y}^{2}=2\left(\frac{1}{2}\right)$

$\Rightarrow \mathrm{y}= \pm 1$

\begin{aligned} \text { Area bounded } & =\int_{-1}^{1}\left(\frac{1}{1+y^{2}}-\frac{y^{2}}{2}\right) d y \\& =\left.\left(\tan ^{-1} y-\frac{y^{3}}{6}\right)\right|_{-1} ^{1} \\& =\frac{\pi}{2}-\frac{1}{3} \end{aligned}

Answer key and solution verified before publishing.

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