Question

Find an equation equivalent to $x^2-y^2=4$ in polar coordinates.

Answer 1

Explanation to correct answer:

Finding the polar coordinates:

Given,

$x^2-y^2=4$

In order to find the polar coordinates,

Substitute $x=r\cos \left(\theta \right)$ and $y=r\sin \left(\theta \right)$

$$\begin{gathered} {\left(r\cos \left(\theta \right)\right)}^2-{\left(r\sin \left(\theta \right)\right)}^2=4 \\ \Rightarrow r^2{\cos }^2\left(\theta \right)-r^2{\sin }^2\left(\theta \right)=4 \\ \Rightarrow r^2\left({\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\right)=4 \\ \Rightarrow r^2\left({\cos }^2\left(\theta \right)-\left(1-{\cos }^2\left(\theta \right)\right)\right)=4\left[\because {\sin }^2\left(\theta \right)=1-{\cos }^2\left(\theta \right)\right] \\ \Rightarrow r^2\left({\cos }^2\left(\theta \right)-1+{\cos }^2\left(\theta \right)\right)=4 \\ \Rightarrow r^2\left(2{\cos }^2\left(\theta \right)-1\right)=4 \\ \Rightarrow r^2=\frac{4}{2{\cos }^2\left(\theta \right)-1} \\ \Rightarrow r=\frac{2}{\sqrt{2{\cos }^2\left(\theta \right)-1}} \end{gathered}$$

substitute $r=\frac{2}{\sqrt{2{\cos }^2\left(\theta \right)-1}}$

$x=\frac{2\cos \left(\theta \right)}{\sqrt{2{\cos }^2\left(\theta \right)-1}}$

$y=\frac{2\sin \left(\theta \right)}{\sqrt{2{\cos }^2\left(\theta \right)-1}}$

Hence, the polar coordinates are $\left(\frac{2\cos \left(\theta \right)}{\sqrt{2{\cos }^2\left(\theta \right)-1}},\frac{2\sin \left(\theta \right)}{\sqrt{2{\cos }^2\left(\theta \right)-1}}\right)$.