Polar Equation to Rectangular?

Question

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.

Answer 1

$$r = \dfrac{4}{1+2 \sin(\theta)} \rightarrow r(1 + 2 \sin(\theta)) = 4$$

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:

$$3y^2 - 16y - x^2 + 16 = 0$$

Answer 2

You can write this as $$4=r+2r\sin\theta=r+2y.$$ Solve for $r$ and square: $$r^2=(4-2y)^2$$ which gives $$x^2=16+3y^2-16y.$$ Now complete the square to bring it into standard form (a hyperbola).

Answer 3

Small fix to other answers. You indeed can do some algebra and get to the point when $$ r + 2r\sin \theta = 4 \implies r = 4 - 2y $$ But then, you need to keep in mind that $r > 0$, so $y < 2$, which is only the lower branch of hyperbola. If you simply square that relation, you take into account spurious solution, that comes from $r = 2y - 4$ relation, which is in turn a parametric form of $$ r = -\frac 4{1 - 2\sin \theta} $$ and that's upper branch equation.