JEE Advanced 2018 — Paper 2 — Question 31

$\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}}$

$\mathrm{y}=\mathrm{mx} \pm \frac{1}{\sqrt{2}} \sqrt{1+\mathrm{m}^{2}} \Rightarrow \frac{2}{\mathrm{~m}^{2}}=1+\mathrm{m}^{2}$

$\Rightarrow \quad\left(\mathrm{m}^{2}+2\right)\left(\mathrm{m}^{2}-1\right)=0$

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

$\Rightarrow y=x+1 \& y=-x-1$

$\Rightarrow \mathrm{Q} \equiv(-1,0)$

$e=\sqrt{1-\frac{1 / 2}{1}}=\frac{1}{\sqrt{2}}$

L.R. $=\frac{2 b^{2}}{a}=\frac{2 \times 1 / 2}{1}=1$

Ellipse $=\frac{x^{2}}{1}+\frac{y^{2}}{1 / 2}=1 \Rightarrow y=-\frac{1}{\sqrt{2}} \sqrt{1-x^{2}}$

$$\begin{aligned} & \text { Area of region }=2 \int_{1 / \sqrt{2}}^{1} \frac{1}{\sqrt{2}} \sqrt{1-\mathrm{x}^{2}} \mathrm{dx} \\& \quad=\sqrt{2}\left[\left|\frac{\mathrm{x}}{2} \sqrt{1-\mathrm{x}^{2}}+\frac{1}{2} \sin ^{-1} \mathrm{x}\right|_{1 / \sqrt{2}}^{1}\right] \\& =\sqrt{2}\left(-\frac{1}{2 \sqrt{2}} \frac{1}{\sqrt{2}}+\frac{\pi}{4}-\frac{\pi}{8}\right) \\& =\sqrt{2}\left(\frac{\pi}{8}-\frac{1}{4}\right)=\frac{(\pi-2)}{4 \sqrt{2}} \end{aligned}$$