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

Question

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...

Answer 1

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 $$ x^2 = 36 + 12y \Longrightarrow y = \frac{x^2}{12} - 3 $$

Answer 2

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

Answer 3

$$r = \frac {6}{1-\sinθ}$$

$${r-r\sinθ}=6$$

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

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

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