# Ellipse centered at a focus

• Oct 2nd 2009, 08:29 AM
synclastica_86
Ellipse centered at a focus
I want to get the polar equation for a ellipse in polar form with the center at one of the foci. I started with the equation of an ellipse in Cartesian coordinates:
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Using $\displaystyle x=rcos\theta$ and $\displaystyle y=rsin\theta$ I obtained:
$\displaystyle \frac{r^2cos^2\theta}{a^2}+\frac{r^2sin^2\theta}{b ^2}=1$
$\displaystyle b^2r^2cos^2\theta+a^2r^2sin^2\theta=a^2b^2$
$\displaystyle r^2(b^2cos^2\theta+a^2sin^2\theta)=a^2b^2$
$\displaystyle r=\frac{ab}{\sqrt{(b^2cos^2\theta+a^2sin^2\theta)} }$

Now I would like the same thing but centered at a focus:
From wiki, I found that the distance between the foci is $\displaystyle \sqrt{a^2-b^2}$. I not sure why that is the case and it would be nice to see some proof. Then I reasoned the ellipse must be shifted half that distance as below:
$\displaystyle \frac{(x-\frac{\sqrt{a^2-b^2}}{2})^2}{a^2}+\frac{y^2}{b^2}=1$
$\displaystyle \frac{x^2+\frac{a^2-b^2}{4}-\sqrt{a^2-b^2}}{a^2}+\frac{y^2}{b^2}=1$
$\displaystyle \frac{x^2}{a^2}+\frac{1}{4}(1-\frac{b^2}{a^2})-\frac{1}{a}\sqrt{1-\frac{b^2}{a^2}}+\frac{y^2}{b^2}=1$
Using the identity $\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$ I obtained:
$\displaystyle \frac{x^2}{a^2}+\frac{e^2}{4}-\frac{e}{a}+\frac{y^2}{b^2}=1$
$\displaystyle \frac{r^2cos^2\theta}{a^2}+\frac{e^2}{4}-\frac{e}{a}+\frac{r^2sin^2\theta}{b^2}=1$
$\displaystyle \frac{r^2cos^2\theta}{a^2}+\frac{r^2sin^2\theta}{b ^2}=\frac{e}{a}-\frac{e^2}{4}$
$\displaystyle r^2\frac{a^2sin^2\theta+b^2cos^2\theta}{a^2b^2}=\f rac{e(4-ae)}{4a}$
$\displaystyle r^2=\frac{e(4-ae)}{4a}\frac{a^2b^2}{a^2sin^2\theta+b^2cos^2\thet a}$
$\displaystyle r^2=\frac{e(4-ae)}{4}\frac{ab^2}{a^2(1-cos^2\theta)+b^2cos^2\theta}$
$\displaystyle r^2=\frac{e(4-ae)}{4}\frac{ab^2}{a^2-(a^2-b^2)cos^2\theta}$
$\displaystyle r^2=\frac{e(4-ae)}{4}\frac{\frac{b^2}{a}}{1-\frac{a^2-b^2}{a^2}cos^2\theta}$
$\displaystyle r^2=\frac{e(4-ae)}{4}\frac{\frac{b^2}{a}}{1-e^2cos^2\theta}$

I'm kindda stuck here. I know that I should get something like $\displaystyle r=\frac{a(1-e^2)}{1+ecos\theta}$
Therefore:
$\displaystyle r^2=\frac{a^2(1-2e^2+e^4)}{1+e^2cos^2\theta+2ecos\theta}$
Where will the $\displaystyle cos\theta$ term came from? and the $\displaystyle e^4$? Am I on the right track?
Am I doing something wrong?
• Oct 2nd 2009, 12:03 PM
Amer
for ellipse use this sub

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

let $\displaystyle \frac{x}{a} = r\cos \theta , \frac{y}{b} = r\sin \theta$ so

$\displaystyle r^2 \cos ^2 \theta + r^2 \sin ^2 \theta =1$

$\displaystyle r^2 = 1$ and do same sub for the second question
• Oct 2nd 2009, 02:34 PM
HallsofIvy
By the way, you do NOT want to get the "center" at one of the foci- that is impossible. You want to have the origin of the coordinate system at one focus.
• Oct 2nd 2009, 04:29 PM
mr fantastic
Quote:

Originally Posted by HallsofIvy
By the way, you do NOT want to get the "center" at one of the foci- that is impossible. You want to have the origin of the coordinate system at one focus.

$\displaystyle \lim_{\text{foci} \rightarrow \text{centre}} \text{Ellipse} = \text{Circle}$.
• Oct 3rd 2009, 01:05 AM
synclastica_86
Quote:

Originally Posted by Amer
for ellipse use this sub

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

let $\displaystyle \frac{x}{a} = r\cos \theta , \frac{y}{b} = r\sin \theta$ so

$\displaystyle r^2 \cos ^2 \theta + r^2 \sin ^2 \theta =1$

$\displaystyle r^2 = 1$ and do same sub for the second question

No I can't do that. That's just a circle with radius = 1. What I want to do is have the equation of an ellipse with the center of the coordinate system centered at one of the foci. From wiki, I know that the answer should be:
$\displaystyle r=\frac{a(1-e^2)}{1+ecos\theta}$ with $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}$

I am actually using this in a physics problem showing that the orbital velocity at the point which the minor axis meets the elliptical orbit is the geometrical mean of the velocities at the ends of the major axis. With this identity: $\displaystyle r=\frac{a(1-e^2)}{1+ecos\theta}$
my proof works very well, but I don't want to just pull an equation off wiki and rest my proof on that. I would like to derive that equation by doing a shift of coordinates.
• Oct 3rd 2009, 02:05 AM
mr fantastic
Quote:

Originally Posted by synclastica_86
No I can't do that. That's just a circle with radius = 1. What I want to do is have the equation of an ellipse with the center of the coordinate system centered at one of the foci. From wiki, I know that the answer should be:
$\displaystyle r=\frac{a(1-e^2)}{1+ecos\theta}$ with $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}$

I am actually using this in a physics problem showing that the orbital velocity at the point which the minor axis meets the elliptical orbit is the geometrical mean of the velocities at the ends of the major axis. With this identity: $\displaystyle r=\frac{a(1-e^2)}{1+ecos\theta}$
my proof works very well, but I don't want to just pull an equation off wiki and rest my proof on that. I would like to derive that equation by doing a shift of coordinates.

The eccentricity of a circle is equal to 1. A circle is an ellipse whose foci are at the centre.