Question

Area enclosed by the curve $\pi\left[4\left(x-\sqrt{2}\right)^{2} +y^{2}\right]=8$ is

• A. $\pi$
• B. $2$
• C. $3\, \pi$
• D. $4$

Answer 1

The Correct Option is D

The given curve is
$\pi\left[4\left(x-\sqrt{2}\right)^{2} +y^{2}\right]=8$
$4\left(x-\sqrt{2}\right)^{2} +y^{2} =\frac{8}{\pi}$
$\left(x-\sqrt{2}\right)^{2} +\left(\frac{y}{2}\right)^{2} =\frac{2}{\pi}$
$\frac{\left(x-\sqrt{2}\right)^{2}}{\left(\sqrt{\frac{2}{\pi}}\right)^{2}}+\frac{y^{2}}{\left(2\sqrt{\frac{2}{\pi }}\right)^{2}}=1\quad\quad...\left(1\right)$
This is the equation of the ellipse having centre ( $\sqrt{ 2}$,0 ).
Observe the figure of ellipse (1). The centre P is ( $\sqrt{ 2}$,0 ). A and B are $\left(\sqrt{2},-\sqrt{\frac{2}{\pi},0}\right)$ and
$\left(\sqrt{2},+\sqrt{\frac{2}{\pi},0}\right)$ respectively.
The required area = 4 ? area of figure PQB
$=4\times\int^{ \sqrt{2}+\sqrt{\frac{2}{\pi}}}_{ \sqrt{2}}$ydx
$=4\times \int_{\sqrt{2}}^{\sqrt{2}+\sqrt{\frac{2}{\pi }}}\sqrt{\frac{8}{\pi}-4\left(x-\sqrt{2}\right)^{2}}$ dx
$=4\times2\int^{\sqrt{2}+\sqrt{\frac{2}{\pi }}}_{\sqrt{2}}\sqrt{\left(\sqrt{\frac{2}{\pi}}\right)^{2}-\left(x-\sqrt{2}\right)^{2}} $dx
$=8\left[\frac{x-\sqrt{2}}{2}\sqrt{\frac{2}{\pi}-\left(x-\sqrt{2}\right)^{2}}+
\frac{\left(2/\pi\right)}{2}sin^{-1}\left(\frac{x-\sqrt{2}}{\sqrt{2/\pi}}\right)\right]_{_{_{_{_{\sqrt{2}}}}}}^{^{\sqrt{2}+\sqrt{\frac{2}{\pi}}}}$
$\left[\because \int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}x\sqrt{a^{2}-x^{2}}+.\frac{a^{2}}{2}sin^{-1}\frac{x}{a}+C\right]$
$=8\left[\begin{Bmatrix}\frac{\sqrt{2}+\sqrt{\frac{2}{\pi}}-\sqrt{2}}{2}\sqrt{\frac{2}{\pi}-\left(\sqrt{2}+\sqrt{\frac{2}{x}-\sqrt{2}}\right)^{2}}+\frac{1}{\pi}sin^{-1}\left(\frac{\sqrt{2}+\sqrt{\frac{2}{\pi}-\sqrt{2}}}{\sqrt{\frac{2}{\pi}}}\right)-\frac{\sqrt{2}-\sqrt{2}}{2}2\sqrt{\frac{2}{\pi}-\left(\sqrt{2}-\sqrt{2}\right)^{2}}+\frac{1}{\pi}sin^{-1}\left(\frac{\sqrt{2}-\sqrt{2}}{\sqrt{2\pi}}\right)\end{Bmatrix}\right]$
$=8\left[\frac{1}{\sqrt{2\pi}}\left(0\right)+\frac{1}{\pi}sin^{-1}\left(1\right)-0-0\right]$
$=8\left(\frac{1}{\pi}\times\frac{\pi}{2}\right)$=4 square units.