The equivalent polar equation for the given rectangular-coordinate equation: x^{2} + y^{2} + 8x=0

Question
Alternate coordinate systems
asked 2020-12-24
The equivalent polar equation for the given rectangular-coordinate equation:
\(\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}\)

Answers (1)

2020-12-25
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:
\(\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}\)
\(\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},\ {\sin{\theta}}={\frac{{{y}}}{{{r}}}},\ {\tan{\theta}}={\frac{{{x}}}{{{y}}}}\)
Converting into equivalent polar coordinates:
\(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}\right.}\)
Put \(\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\)
\(\displaystyle\Rightarrow\ {\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\ +\ {8}{r}{\cos{\theta}}={0}\)
\(\displaystyle\Rightarrow\ {r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}\ +\ {8}{r}{\cos{\theta}}={0}\)
\(\displaystyle\Rightarrow\ {r}^{{{2}}}\ {\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\ }\theta\right)}\ +\ {8}{r}{\cos{\theta}}={0}\)
\(\displaystyle\le{f}{t}{\left\lbrace{{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\ }\theta={0}{r}{i}{g}{h}{t}\right\rbrace}\)
\(\displaystyle\Rightarrow\ {r}^{{{2}}}\ +\ {8}{r}{\cos{\theta}}={0}\)
\(\displaystyle\Rightarrow\ {r}\ +\ {8}{\cos{\theta}}={0}\)
Hence, desired equivalent polar coordinates would be \(\displaystyle{r}\ +\ {8}{\cos{\theta}}={0}\)
0

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