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Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$

The answer should be: $y = \frac{1}{12} x^2 -3$

But how to arrive at the answer?

I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.

I also ended up with $r = \frac{6}{\frac{r-y}{r}} $ -> $ 1 = \frac{6}{\sqrt{x^2+y^2}-y}$ -> x = +- 6

but that's not the answer...

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Just multiply by denominator: $$ r = \frac{6}{1-\sin\theta} \Longrightarrow r(1 - \sin\theta) = 6\Longrightarrow r - y = 6 $$ Now $$ \sqrt{x^2+y^2} = 6+y\Longrightarrow x^2 + y^2 = (6+y)^2, $$ or $$ x^2 = 36 + 12y \Longrightarrow y = \frac{x^2}{12} - 3 $$

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  • $\begingroup$ @Jack, $r - y = 6\Longrightarrow r = y+6$, isn'i it? $\endgroup$ – Michael Galuza Jul 1 '15 at 3:42
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    $\begingroup$ @Jack, we divide $36$ by $12$: $12y=x^2-36$, $y=x^2/12-36/12=x^2/12-3$ $\endgroup$ – Michael Galuza Jul 1 '15 at 3:55
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$$r = \frac {6}{1-\sinθ}$$

$${r-r\sinθ}=6$$

$$\sqrt{x^2 + y^2}-y=6$$

$$\sqrt{x^2 + y^2}=y+6$$ now squaring both side and simplyfying we get

$$y = \frac{1}{12} x^2 -3$$

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Write $$6=r(1-\sin\theta)=r-r\sin\theta=\sqrt{x^2+y^2}-y$$ Then $$6+y=\sqrt{x^2+y^2}\implies12y+36=x^2$$

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