Abril Huynh

2023-03-26

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

unnlattpdui

Beginner2023-03-27Added 5 answers

Explanation to correct answer:

How to determine the polar coordinates:

We have,

${x}^{2}-{y}^{2}=4$

The polar coordinates can be discovered using,

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

${\left(r\mathrm{cos}\left(\theta \right)\right)}^{2}-{\left(r\mathrm{sin}\left(\theta \right)\right)}^{2}=4\Rightarrow {r}^{2}{\mathrm{cos}}^{2}\left(\theta \right)-{r}^{2}{\mathrm{sin}}^{2}\left(\theta \right)=4\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-{\mathrm{sin}}^{2}\left(\theta \right)\right)=4\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-\left(1-{\mathrm{cos}}^{2}\left(\theta \right)\right)\right)=4\left(\because {\mathrm{sin}}^{2}\left(\theta \right)=1-{\mathrm{cos}}^{2}\left(\theta \right)\right)\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-1+{\mathrm{cos}}^{2}\left(\theta \right)\right)=4\Rightarrow {r}^{2}\left(2{\mathrm{cos}}^{2}\left(\theta \right)-1\right)=4\Rightarrow {r}^{2}=\frac{4}{2{\mathrm{cos}}^{2}\left(\theta \right)-1}\Rightarrow r=\frac{2}{2}$

substitute $r=\frac{2}{2}$

$x=\frac{2\mathrm{cos}\left(\theta \right)}{2}$

$y=\frac{2\mathrm{sin}\left(\theta \right)}{2}$

Therefore, the polar coordinates are $\left(\frac{2\mathrm{cos}\left(\theta \right)}{2},\frac{2\mathrm{sin}\left(\theta \right)}{2}\right)$.

How to determine the polar coordinates:

We have,

${x}^{2}-{y}^{2}=4$

The polar coordinates can be discovered using,

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

${\left(r\mathrm{cos}\left(\theta \right)\right)}^{2}-{\left(r\mathrm{sin}\left(\theta \right)\right)}^{2}=4\Rightarrow {r}^{2}{\mathrm{cos}}^{2}\left(\theta \right)-{r}^{2}{\mathrm{sin}}^{2}\left(\theta \right)=4\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-{\mathrm{sin}}^{2}\left(\theta \right)\right)=4\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-\left(1-{\mathrm{cos}}^{2}\left(\theta \right)\right)\right)=4\left(\because {\mathrm{sin}}^{2}\left(\theta \right)=1-{\mathrm{cos}}^{2}\left(\theta \right)\right)\Rightarrow {r}^{2}\left({\mathrm{cos}}^{2}\left(\theta \right)-1+{\mathrm{cos}}^{2}\left(\theta \right)\right)=4\Rightarrow {r}^{2}\left(2{\mathrm{cos}}^{2}\left(\theta \right)-1\right)=4\Rightarrow {r}^{2}=\frac{4}{2{\mathrm{cos}}^{2}\left(\theta \right)-1}\Rightarrow r=\frac{2}{2}$

substitute $r=\frac{2}{2}$

$x=\frac{2\mathrm{cos}\left(\theta \right)}{2}$

$y=\frac{2\mathrm{sin}\left(\theta \right)}{2}$

Therefore, the polar coordinates are $\left(\frac{2\mathrm{cos}\left(\theta \right)}{2},\frac{2\mathrm{sin}\left(\theta \right)}{2}\right)$.

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