Question

The polar coordinates of the point whose Cartesian coordinates are $(-\frac{1}{\sqrt{2}},\sqrt{2})$, are

• A. $(\sqrt{\frac{5}{2}},\pi -{\tan }^{-1}(2\left)\right)$
• B. $(\sqrt{\frac{5}{2}},-{\tan }^{-1}(2\left)\right)$
• C. $(\sqrt{\frac{5}{2}},{\tan }^{-1}(2\left)\right)$
• D. $(\sqrt{\frac{5}{2}},\pi +{\tan }^{-1}(2\left)\right)$

Answer 1

The correct option is A $(\sqrt{\frac{5}{2}},\pi -{\tan }^{-1}(2\left)\right)$

Let $(x,y)\equiv (-\frac{1}{\sqrt{2}},\sqrt{2})$
Since, $r=\sqrt{x^2+y^2}$
$\Rightarrow r=\sqrt{{(-\frac{1}{\sqrt{2}})}^2+(\sqrt{2}{)}^2}\Rightarrow r=\sqrt{\frac{1}{2}+2}=\sqrt{\frac{5}{2}}$

Also, $(x,y)\equiv (-\frac{1}{\sqrt{2}},\sqrt{2})$ lies in $2^{\text{nd}}$ quadrant.
$\therefore \theta =\pi -{\tan }^{-1}\left|\frac{y}{x}\right|\Rightarrow \theta =\pi -{\tan }^{-1}|\sqrt{2}\times (-\frac{\sqrt{2}}{1})|\Rightarrow \theta =\pi -{\tan }^{-1}\left(2\right)\therefore (r,\theta )\equiv (\sqrt{\frac{5}{2}},\pi -{\tan }^{-1}(2\left)\right)$