<s>[INST] Polar Equation to Rectangular? The equation is:
$$r = \frac{4}{1+2sin(\theta)}$$
I'm confused about how to convert it into rectangular form.
This is what I have so far, although I'm not sure it's correct:
$$r = \frac{ 4(1-2sin(\theta)) }{ 1-2sin^2(\theta) }  $$
$$r = \frac{ 4-8sin(\theta) } {( cos^2(\theta) - sin^2(\theta) )}  $$
$$r = \frac{( 4-8(\frac{y}{r}) )}{ ( (\frac{x}{r})^2 - (\frac{y}{r})^2 )}  $$
$$r ( (\frac{x}{r})^2 - (\frac{y}{r})^2 ) = 4 - 8(\frac{y}{r})  $$
$$\frac{(x^2)}{r} - \frac{(y^2)}{r} - \frac{8y}{r} = 4$$
What do I do after that? Do I complete the square? That leaves me with an r on the right-hand side, though. 
 [/INST] $$r = \dfrac{4}{1+2 \sin(\theta)} \rightarrow r(1 + 2 \sin(\theta)) = 4$$
A polar plot shows:

We have:


*

*$r = \sqrt{x^2 + y^2}$

*$x = r \cos(\theta)$

*$y = r \sin(\theta)$


So,
$$r(1 + 2 \sin(\theta)) = r + 2 r \sin(\theta) = \sqrt{x^2 + y^2} + 2 y = 4$$
We can re-write this as:
$$3y^2 - 16y - x^2 + 16 = 0$$
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