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In finding polar equations of conics we take focus as the origin but ellipse has 2 focci so which focus is taken as the origin?
Or If if focus is not taken on origin then how to find the polar equation of ellipse?

user606262
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In the simplest polar equation for the ellipse, take one focus at the origin and the second focus as $r=b,\theta=0$ with$b>0$. Thus the second focus would lie along the positive $x$ axis in rectangular coordinates or the ray $\theta=0$ in polar coordinates.

Let $O$ be the origin, $A$ be the second focus at $r=b>0,\theta=0$ and $P$ be a point on the ellipse. Defining $e$ as the eccentricity of the ellipse we then have from the Law of Cosines on $\triangle OAP$:

$|AP|^2=|OA|^2+|OP|^2-2|OA||OP|\cos \theta$

$((b/e)-r)^2=b^2+r^2-2br\cos \theta$

We expand the left side and note that the quadratic terms in $r$ cancel out leading to an equation that is linear in $r$:

$(b/e)^2-b^2=2br((1/e)-\cos \theta)$

Solving for $r$:

$r=\frac{b(1-e^2)/2e}{1-e \cos \theta}=\frac{a}{1-e \cos \theta}$

where $a$ is the semilatus rectum of the ellipse.

Oscar Lanzi
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    +1. It may be worth noting that the "simplest" equation is actually a tie: $$r = \frac{a}{1-e\cos\theta}\qquad\text{vs}\qquad r=\frac{a}{1+e\cos\theta}$$ Both have a focus at the origin. As you mention, the first has its other focus on the positive $x$-axis; the second has it on the negative $x$-axis. I prefer the latter, so that, as $e$ approaches $1$ and the conic approaches a parabola, the figure gets wider off to the left, while the behavior at $\theta=0$ (the traditional "start" of an angular sweep) remains comparatively steady. (That steadiness persists for hyperbolic sweeps with $e>1$.) – Blue Oct 20 '18 at 10:22
  • How to trace the conic by this equation? – user606262 Oct 20 '18 at 10:26
  • Also, readers should be aware that your use of $a$ and $b$ (and thus also my use in the above comment) is different than is typically used to describe ellipses. – Blue Oct 20 '18 at 10:29
  • @blue I know the nomenclature is different. But the form I first saw used $a$ as the semilatus rectum leading me down the garden path. – Oscar Lanzi Oct 20 '18 at 10:51