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I am trying to approximate the parametric equation of an ellipse with one focus at $(0,0)$. The semi-major axis, $a$, and the eccentricity, $e_c$, are known variables, with $0 \le e_c < 1$, and $a \neq 0$. I have a rough approximation of the parametric: $$\left(na^{2}\left(1-e_{c}\right)\left(\left(e_{c}-1\right)\left(e_{c}+1\right)a\right)\left(\cos\left(t\right)\right)-a\left(e_{c}\right),m\left(e_{c}+1\right)\left(1-e_{c}\right)\left(\left(e_{c}-1\right)\left(e_{c}+1\right)a\right)\cdot\sin\left(t\right)\right)$$

$m$ and $n$ are constants in terms of $a$ and $e_c$. So, what are $m$ and $n$, and if this form doesn't work, then what is the proper form? If you need reference, here is my graph.

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You can use the usual parametric equation and shift $x$ to bring a focus to the origin:

$$\begin{cases}x=a\cos t-f,\\y=b\sin t\end{cases}$$ where $f$ is the half focus distance ($b=a\sqrt{1-e^2}, f=ea$).

You can also use the polar equation

$$\rho=\frac p{1-e\cos\theta}$$ ($p=\dfrac{(1-e^2)a}e$) and $$\begin{cases}x=\rho\cos\theta,\\y=\rho\sin\theta.\end{cases}$$

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  • $\begingroup$ Thanks, but would we have $x = a \cos t \pm f$ instead of having a $+a$ in it? $\endgroup$ – fasterthanlight Oct 18 '20 at 22:08
  • $\begingroup$ @FTL'saccountwillbedeleted: ooops, yes. $\endgroup$ – Yves Daoust Oct 19 '20 at 7:10

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