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Wikipedia gives the following equation for the conic sections in the polar coordinate system:

$r = \frac{l}{1+e\cos\varphi}$.

According to the article on conic sections, in case of an ellipse $e = \sqrt{1-\frac{b^2}{a^2}}$ and $l = \frac{b^2}{a}$, where $a$ is the semi-major axis and $b$ is the semi-minor axis.

However, when I try to substitute concrete values into the equation, I don't get the results I expect. For example, let $a = 2$ and $b = 1$. It seems clear to me that for $\varphi = 0$ I should get the rightmost point of the ellipse, that is $r = a = 2$, correspoding to $(a, 0) = (2, 0)$ in the Cartesian coordinates. However:

$l = \frac{b^2}{a} = \frac{1}{2}$

$e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1 - \frac{1^2}{2^2}} = \frac{\sqrt{3}}{2}$

$r = \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{1}{2 + \sqrt{3}}$.

It's hardly two. I am sorry if I am making a fool of myself, but where do I make a mistake?

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The polar center for your equation is a focus of the ellipse, not its center.

Polar form relative to the center of the ellipse is $$r(\phi)=\frac{ab}{\sqrt{(a\sin\phi)^2+(b\cos\phi)^2}}$$

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what you calculated was the distance to one of the focals. if you calculate $r(\varphi=\pi)=\frac{1}{2- \sqrt{3}}$ you may notice that they add up to $4=2a$.

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