Parametric equation of ellipse with foci at origin I want to know what the parametric equation for an ellipse is if the one of the foci is centered at the origin. I know the semi-major and minor axes. 
I know the parametric equation of an ellipse where the center of the ellipse is at the origin is $x = a\cos(t)$ and $y = b\sin(t)$ but I want to know the parametric equation where one of the foci is at the origin.
 A: In polar coordinates, the equation for an ellipse with one focus at the origin, and whose center lies in the direction $\phi$ from the origin, is
$$r = \frac{a(1-e^2)}{1 - e\, \cos(\theta - \phi)}$$
where $a$ is the semi-major axis, $b$ is the semi-minor axis, and
$e = \sqrt{1 - \frac{b^2}{a^2}}$. 
Using typical data available about planetary orbits, you can set
$a$ equal to the planet's mean distance from the Sun
and $e$ equal to the eccentricity of the orbit.
To parameterize the $x$ and $y$ coordinates,
just convert from polar to Cartesian coordinates: $x = r \cos \theta$,
$y = r \sin \theta$.
An advantage of this formulation is you can plot the perihelion of the orbit in any
direction you want, and you can easily plot the orbits of multiple planets
(each of which has its perihelion in a different direction from the Sun).
A: It's easiest to just shift the ellipse. Assuming your major axis is horizontal, your set of equations will be $x(t)=a\cos (t) + \sqrt{a^2-b^2}$ and $y(t)=b\cos (t)$. The $\sqrt{a^2-b^2}=c$ is the distance from the focus to the center.
I have found, however, that when trying to simulate planetary motion a polar equation is a better alternative so long as you have a graphing utility that can handle polar curves. This is better because of Kepler's second law, which states that equal areas are swept out by the planet in equal amounts of time. The equal areas can best be calculated by a polar integral, and then a predetermined time step can be set equal to this integral. I'm still working on a full solution of a polar equation with a focus at the origin. The following URL has what I've just whipped up with the center of the ellipse at the origin. https://www.desmos.com/calculator/vfrxnqo5dy
Note that both equations display the same graph. The polar equation could be simplified more.
EDIT: Seems like the crux of the situation is trying to solve $\tan(\theta) = \frac{b\sin (t)}{a\cos (t) + \sqrt{a^2-b^2}}$ for $t$. I'll keep working.
