How would I convert $2x^2 + y^2 + 3y = 0$ into polar form?

Question

We are currently working with rectangular and polar equations.

How would I convert $$2x^2 + y^2 + 3y = 0$$ into polar form?

So far, I have tried to make the equation into rectangular and back into polar, but I am having no luck as I cannot find a way to get rid of the $2$ coefficient.

All help is appreciated!

Answer 1

We have $$2x^2+y^2+3y = \color{red}{x^2}+\color{blue}{x^2+y^2}+\color{green}{3y} = \color{red}{r^2\cos^2\theta} + \color{blue}{r^2} + \color{green}{3r\sin\theta} = 0,$$so assuming $r \neq 0$, we get: $$r = \frac{-3\sin\theta}{1+\cos^2\theta}.$$

Answer 2

Write

$$2x^2+y^2+3y = x^2 +(x^2+y^2)+3y$$ and use $$x^2+y^2 = r^2,\quad x = r\cos(\theta),\quad y = r\sin(\theta)$$ and solve for $r$.

Hint: consider the case $r=0$ separately.

Answer 3

HINT: you will need the equations $$x=r\cos(\theta),y=r\sin(\theta)$$ with $$r=\sqrt{x^2+y^2}$$

Answer 4

I am not sure what you are intending to do, but if it is the usual polar coordinates, using $x=r\cos\theta, y=r\sin\theta$, with $x^{2}+y^{2}=r^{2}$, we have

$2x^{2}+y^{2}+3y=2r^{2}\cos^{2}\theta+r^{2}\sin^{2}\theta+3r\sin\theta=r^{2}\cos^{2}\theta+r^{2}+3r\sin\theta$

Or if you prefer,

$r^{2}\cos^{2}\theta+r^{2}+3r\sin\theta=r^{2}(1-\sin^{2}\theta)+r^{2}+3r\sin\theta=2r^{2}-r^{2}\sin^{2}\theta+3r\sin\theta$

Not sure if this is what you are looking for.