Question

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
\(\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}\)

Answer 1

Concept used: Conversion formula for coordinate systems are given as
a)
From polar to rectangular:
\(\displaystyle{x}={r} \cos{\theta}\)
\(\displaystyle{y}={r} \sin{\theta}\)
b)
From rectangular to polar:
\(\displaystyle{r}=\pm\sqrt{{{x}^{2}+{y}^{2}}}\)
\(\displaystyle \cos{\theta}=\frac{x}{{r}}; \sin{\theta}=\frac{y}{{r}}; \tan{\theta}=\frac{x}{{y}}\)
Calculation:
Converting into equivalent polar equation
\(\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}\)
Put \({\left\lbrace{x}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\cos{{\left\lbrace\theta\right\rbrace}}},{\left\lbrace{y}\right\rbrace}={\left\lbrace{r}\right\rbrace}{\sin{{\left\lbrace\theta\right\rbrace}}},\)
\(\displaystyle{x}={r} \cos{\theta},{y}={r} \sin{\theta}\)
\(\displaystyle\Rightarrow{\left({r} \cos{\theta}\right)}^{2}+{\left({r} \sin{\theta}\right)}^{2}+{8}{r} \cos{\theta}={0}\)
\(\displaystyle\Rightarrow{r}^{2}{{\cos}^{2}\theta}+{r}^{2}{{\sin}^{2}\theta}+{8}{r} \cos{\theta}={0}\)
\(\displaystyle\Rightarrow{r}^{2}{\left({{\cos}^{2}\theta}+{{\sin}^{2}\theta}\right)}+{8}{r} \cos{\theta}={0}{\left\lbrace{{\cos}^{2}\theta}+{{\sin}^{2}\theta}={1}\right\rbrace}\)
\(\displaystyle\Rightarrow{r}^{2}+{8}{r} \cos{\theta}={0}\)
\(\displaystyle\Rightarrow{r}+{8} \cos{\theta}={0}\)
Thus, desired equivalent polar equation would be \(\displaystyle{r}+{8} \cos{\theta}={0}\)