polar coordinates
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!
The other equation you need is $\theta = \arctan \frac {y}{x}$ and you have to worry about the branches of arctan. So now you have $\sqrt{x^2+y^2}=\frac{1}{20} \arctan \frac {y}{x}$. To translate right and down, you replace $x$ with $x+a$ and $y$ with $y-b$. And you have quite a mess. Taking $\sqrt{(x+a)^2+(y-b)^2}=\frac{1}{20} \arctan \frac {y-b}{x+a}$ back to polar is a lot of work and probably not illuminating.
what is the rectangular form?
It is the $y=f(x)$ expression of the curve in the $x,y$ referential (see picture). It can also be the implicit form $F(x,y)=F(x,f(x))\equiv 0$.
Steps:
1) transformation of polar into rectangular coordinates (also known as Cartesian coordinates) (see picture)
$$x=r\cos \theta ,$$
$$y=r\sin \theta ;$$
2) from trigonometry and from 1) $r=\sqrt{x^2+y^2}$
$$\sin \theta =\frac{y}{r}=\frac{y}{\sqrt{ x^{2}+y^{2}}};$$
3) substitution in the given equation $$r=\frac{9}{4+\sin \theta }=\dfrac{9}{4+\dfrac{y}{\sqrt{x^{2}+y^{2}}}}=9\dfrac{\sqrt{x^{2}+y^{2}}}{4\sqrt{x^{2}+y^{2}}+y};$$
4) from 1) $r=\sqrt{x^2+y^2}$, equate
$$9\frac{\sqrt{x^{2}+y^{2}}}{4\sqrt{ x^{2}+y^{2}}+y}=\sqrt{x^{2}+y^{2}};$$
5) simplify to obtain the implicit equation
$$4\sqrt{x^{2}+y^{2}}+y-9=0;$$
6) Rewrite it as $$4\sqrt{x^{2}+y^{2}}=9-y,$$ square it (which may introduce extraneous solutions, also in this question), rearrange as
$$16y^{2}+18y+15x^{2}-81=0,$$
and solve for $y$
$$y=-\frac{3}{5}\pm \frac{4}{15}\sqrt{81-15x^{2}}.$$
7) Check for extraneous solutions.
if I get the expression in rectangular form, how am I able to convert it back to polar coordinate?
The transformation of rectangular to polar coordinates is
$$r=\sqrt{x^{2}+y^{2}}, \qquad \theta =\arctan \frac{y}{x}\qquad \text{in the first quadrant},$$
or rather $\theta =\arctan2(y,x)$ to take into account a location different from the first quadrant. (Wikipedia link).
As commented by J.M. the curve is an ellipse. Here is the plot I got using the equation $16y^{2}+18y+15x^{2}-81=0$.
Best Answer
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: $$r = \frac{-3\sin\theta}{1+\cos^2\theta}.$$