Polar Equations of Ellipse

Question

I currently have the equation: $$1=\frac{x^2}{75}+\frac{y^2}{56.25}$$
for my ellipse and I have found the values such as the foci($2.5\sqrt{3}, 0$) and ($-2.5\sqrt{3}, 0$) with the eccentricity of 0.5. I have translated this into the form of $r=\frac{ep}{1-ecos(x)}$ $\Rightarrow r=\frac{3.75}{1-0.5cos(x)}$
I understand that this is the correct shape for my ellipse but it does not have the original center at the origin, which is the reason to why I am asking this question. How can I show algebraically rearranging my Cartesian formula to polar formula and still maintain my center at the origin?

Answer 1

To pass to polar coordinates use the change of variable $x=r\cos(t)$, $y=r\sin(t)$.

The equation of the ellipse becomes

$$1=r^2\left(\cos^2(t)/75+\sin^2(t)/56.25\right)$$

Answer 2

Given the ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

there are several ways to express it in the complex plane. Some are shown in the other answers. Here are two direct expressions for $z(\theta),~\theta\in[0,2\pi]$.

$$ z=a\cos\theta+ib\sin\theta $$ $$ r=\frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}} $$ $$ z=re^{i\theta} $$

P.S. I gave the same answer to your about how to express this particular ellipse in Cartesian coordinates. However, let me add here that in the first form $\theta$ is a parameter, whereas in the second form it is actually the polar angle.