<s>[INST] Converting a polar equation to a rectangular one $$r=\frac { 4 }{ 1+2\sin\theta  } $$
Steps I took:
$$(1+2\sin\theta )r=\frac { 4 }{ 1+2\sin\theta  } (1+2\sin\theta )$$
$$r+2r\sin\theta =4$$
$$r+2y=4$$
$$(r+2y)^2=16$$
$$(r+2y)(r+2y)=r^2+4yr+4y^2$$
$$r^2+4yr+4y^2-16=0$$
My outcome doesn't seem to match the correct answer. Where did I go wrong?
 [/INST] use that $$r=\sqrt{x^2+y^2}$$ and $$\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$$ and you will get
$$\sqrt{x^2+y^2}=\frac{4}{1+\frac{2y}{\sqrt{x^2+y^2}}}$$
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