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Convert the polar equation to rectangular form (rectangular equation)

$$r=\frac{9}{1-3\cos(\theta)}$$

I know that $r^2= x^2+y^2, x= r\cos(\theta)$ and $y= r\sin(\theta)$ and $\tan(\theta)= y/x$

I don't even understand how to get started.

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    $\begingroup$ do you mean $r=9-3\cos(\theta)$ or $r=\frac{9}{1-3\cos(\theta)}$? $\endgroup$
    – cirpis
    Jul 16, 2014 at 19:26
  • $\begingroup$ the second one, it is division, sorry I am new to this. $\endgroup$
    – user164756
    Jul 16, 2014 at 19:29

1 Answer 1

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Assuming $$r=\frac{9}{1-3\cos(\theta)}$$ then we can multiply by $1-3\cos(\theta)$ to get $$r-3r\cos(\theta)=9$$ now substituting $r=\sqrt{x^2+y^2}$ and $r\cos(\theta)=x$ to get $$\sqrt{x^2+y^2}-3x=9$$ $$\sqrt{x^2+y^2}=9+3x=3(3+x)$$ square both sides

$$x^2+y^2=3^2(3+x)^2=9(x^2+6x+9)$$ $$y^2=8x^2+54x+81$$ thus we gain that $$y=\pm \sqrt{8x^2+54x+81}$$

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  • $\begingroup$ Thank you! I was getting 8x^2 and 81, but was missing the middle term. My first step was wrong because I wasn't multiplying first. I can see where I went wrong now. I rearranged the answer to equal zero for my homework. $\endgroup$
    – user164756
    Jul 16, 2014 at 19:56

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