I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants $a$ and $b$ in this format - image

Where $2a$ is the total height of the ellipse and $2b$ being the total width. You can then find the radial length, $r$, at any angle $\theta$ to major axis as...

$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$

...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:

$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$

Here is the link to the Wikipedia page: Can someone explain this, please? Why divide by the hypotenuse? Why the $ab$? Thank you!

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    $\begingroup$ The point $(b\cos\theta,a\sin\theta)$ is not at angle $\theta$. $\endgroup$
    – user856
    Commented Feb 27, 2013 at 0:18
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    $\begingroup$ How is it not?? I'm pretty sure it's at angle $\theta$ moving ccw from $y = 0$, $x = b$ EDIT: Oh shit you're right!!! $\theta$ changes at the constant rate of a circle, not at the rate of an ellipse! Thank you!!!! $\endgroup$ Commented Feb 27, 2013 at 0:42
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    $\begingroup$ You may also want to look at my answer to math.stackexchange.com/questions/493104/… $\endgroup$
    – user2469
    Commented Feb 23, 2014 at 18:04
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    $\begingroup$ According to Wolfram Alpha, your version - because of the difference in theta as described, gives an interesting shape that looks like a slightly distorted ellipse: wolframalpha.com/input/… $\endgroup$
    – user148686
    Commented May 8, 2014 at 12:29
  • $\begingroup$ @ Athan Clark: Your typo corrected.Also no division by hypotenuse in the second $ r(\theta)$.Hope ok. $\endgroup$
    – Narasimham
    Commented May 9, 2020 at 16:18

5 Answers 5


It's easiest to start with the equation for the ellipse in rectangular coordinates:

$$(x/a)^2 + (y/b)^2 = 1$$

Then substitute $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$ and solve for $r(\theta)$.

That will give you the equation you found on Wikipedia.


Polar Equation from the Center of the Ellipse

The equation of an ellipse is $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag1 $$ Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(1)$, we get $$ r^2\cos^2(\theta)+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag2 $$ and we can solve $(2)$ for $r^2$ to get the polar equation $$ r^2=\frac{\overbrace{a^2\!\left(1-e^2\right)}^{b^2}}{1-e^2\cos^2(\theta)}\tag3 $$ enter image description here

Polar Equation from a Focus of the Ellipse

Centered at the right focus $$ \left(\frac{x+ae}a\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag4 $$ Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(4)$, we get $$ r^2\cos^2(\theta)+2aer\cos(\theta)+a^2e^2+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag5 $$ which gives the quadratic equation in $r$: $$ \frac{r^2\left(1-e^2\cos^2(\theta)\right)}{1-e^2}+2aer\cos(\theta)-a^2\!\left(1-e^2\right)=0\tag6 $$ whose solution is $$ r=\frac{a\!\left(1-e^2\right)}{1+e\cos(\theta)}\tag7 $$ enter image description here

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    $\begingroup$ This is a very handy and instructive answer, thank you! I've quoted it here. $\endgroup$
    – uhoh
    Commented May 16, 2020 at 15:26


What you at first proposed as ellipse looks like:

enter image description here

The Ellipse parametrization is done differently. To more clearly distinguish between them we should note there are two different $\theta$ s, viz $\theta_{deLaHire}$ and the standard polar coordinate $\theta_{polar}$ used for central conics, ellipse in this case. We are not referring to the Newton Ellipse as there is no query about it.

The first angle denotes by $ \theta_{deLaHire}$.

A radial line was constructed by deLaHire originally commences at a slightly bigger angle $\theta_{\text{deLaHire}};$ (red lines) each point $E$ on ellipse in first quadrant is reached by drawing vertical and horizontal lines from points of intersection of this polar/radial line with the two circles radii $(a,b)$ at $(P,Q)$, to meet at E as shown.

$$ x= a \cos\theta_{deLaHire}\; ; y=b \sin\theta_{deLaHire}\;;\tag1 $$

The second angle is used in polar coordinates in the standard ellipse and is measured from center of the circles. We call it $\theta_{polar}$ as usual. Green radial line.

For an ellipse axes $(a,b)$ along $(x,y)$ coordinate axes respectively centered at origin given Wiki expression is obtained in polar coordinates thus:

Plug in

$$ x=r_{polar}\cos \theta_{polar};\, y=r_{polar}\sin \theta_{polar} ; $$

casting the standard equation of an ellipse from Cartesian form:

$$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 =1 $$

to get

$$ OE =r_{polar}= \frac{ab}{\sqrt{(b \cos \theta_{polar})^2 + (a \sin \theta_{polar})^2}} \tag 2 $$

deLaHire and polar EllipseCoords

In either case polar angles $\theta = 0$ and $\theta= \pi/2$ reach to the same points at the ends of major and minor axes respectively. The angle variations are plotted showing by comparison that starting deLaHire polar line is inclined more than (or equals to at extreme axes) the Central polar coordinate always. Can you figure out errors in the second and fourth quadrants?

Sketched Ellipse dimensions are $(a=5,b=3,e=0.8)$.


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    $\begingroup$ Are the labels in that diagram the right way round? Surely the eccentric anomaly is angle between OP and the x-axis? It is the polar angle that measures the angle of the point on the ellipse from the centre. $\endgroup$ Commented May 8, 2021 at 8:39
  • $\begingroup$ Terribly sorry about this. I made a fresh simplified write-up / answer. Thanks for pointing out. $\endgroup$
    – Narasimham
    Commented Jul 29, 2022 at 5:06

You're making the common mistake of using the polar coordinate instead of the eccentric anomaly which is the parameter in the ellipse coordinates.

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    $\begingroup$ Would you be willing to flesh out more detail, supplying definitions for your vocabulary and clear implication of your reasoning? Just so this answer is self-contained. $\endgroup$ Commented Dec 28, 2016 at 1:06

I write the polar form of ellipse equation in any arbitrary location at this link https://www.desmos.com/calculator/gkijxayubk. The proposed polar formula covers any transformation of an ellipse curve, including the translation, reflection, rotation about the ellipse’s centre, and rotation about the origin without shape distortion. The transformed curve remains its original shape. I also published the curve fitting of an ellipse polar form in this publication https://doi.org/10.3390/f14061102. The formula is in the attached picture.

general polar formula for an ellipse


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