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I am unclear about how to parameterize for x,y on the ellipse shown in the figure based on lines drawn from the focus, point "P", (not the center, point "O") and for the angle $\phi$. I believe the drawing shows what I am looking for. For the standard form of this ellipse, $a=\frac{R}{\sin\beta}$ and b = R.

Edit #1: Removed confusing statement regarding q.

Edit #2: For a standard ellipse the parameterization would be $x = a \cos \alpha$ and $y = b\sin \alpha$. If it is offset to the focus as shown in the diagram, then I think the parameterization is $x = a \cos \alpha + c$ and $y = b\sin \alpha$. The problem seems to be getting the parameterization in terms of $\phi$. Note that the diagram has been revised to show the angle $\alpha$.

enter image description here

rdemo
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  • Your question is a little confusing. What variables are you using to parameterize the ellipse? Also, it might help to answer your question better if you show your attempt at the question. – Vector Feb 10 '24 at 16:54
  • The parameter $\alpha$ in the parametric equation IS NOT the same as the angle in the figure. For the polar equation you need, see here: https://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_focus – Intelligenti pauca Feb 10 '24 at 17:08
  • @intelligenti: If I understand your comment correctly, you are correct. As shown in the diagram $\alpha$ is not the same as $\phi$. I simply showed what I think is the parameterization in terms of $\alpha$. As I stated, I would like to get the parameterization in terms of $\phi$. – rdemo Feb 10 '24 at 17:11
  • @intelligenti: Yes, that looks like it. Thanks. – rdemo Feb 10 '24 at 17:17
  • From the link provided by intelligenti, the equation for q in my drawing is: $q(\phi) = \frac{a(1-e^2)}{(1 \pm e\cos\phi)}$. If I were to use this equation in an integral, would someone confirm that what I need to do is to split up the integral into two parts. One part would be where $\pm = +$ and the second split of the integral would be where $\pm = -$. – rdemo Feb 10 '24 at 17:21
  • You don't have to split anything: if $\phi=0$ points to the far vertex, as in your diagram, then use the minus sign. – Intelligenti pauca Feb 10 '24 at 18:02
  • BTW: my remark on $\alpha$ was related to the standard parameterization centered at $O$. – Intelligenti pauca Feb 10 '24 at 18:04

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