Question

Rectangular Coordinates to Polar Coordinates Convert the rectangular coordinates to polar coordinates with$r>0$ and$0\le \theta <2\pi .$$(-\sqrt{6},-\sqrt{2})$.

Short answer

The polar coordinates of the point $(-\sqrt{6},-\sqrt{2})$are.$(2\sqrt{2},\frac{7\pi }{6})$

Step-by-step solution

Step 1.Given information.

The polar coordinates of the point are$(-\sqrt{6},-\sqrt{2})$ .

Step 2. Since the rectangular coordinates of the point are(−6,−2)  we can say that .x=−6,y=−2

Substituting $(x,y)=(-\sqrt{6},-\sqrt{2})$ in the polar coordinates formulae, we get,

$\begin{array}{c}{r}^2=x^2+y^2\\ \Rightarrow r^2={(-\sqrt{6})}^2+{(-\sqrt{2})}^2\\ \Rightarrow r^2=6+2\\ \Rightarrow r^2=8\\ \Rightarrow r=\pm 2\sqrt{2}\end{array}$

Step3. Finding the value of angle

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{-\sqrt{2}}{-\sqrt{6}}\right)\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ \Rightarrow \theta =\frac{\pi }{6},\frac{7\pi }{6}\end{array}$

Since $r>0$, we consider only the solution, $r=2\sqrt{2}$and this occurs at $\theta =\frac{7\pi }{6}$.