# Ellipse in polar coordinates

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

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 degree, $\theta$, as...

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

...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: http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center Can someone explain this, please? Why divide by the hypotenuse? Why the $ab$? Thank you!

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The point $(b\cos\theta,a\sin\theta)$ is not at angle $\theta$. – Rahul Feb 27 '13 at 0:18
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!!!! – Athan Clark Feb 27 '13 at 0:42
You may also want to look at my answer to math.stackexchange.com/questions/493104/… – barrycarter Feb 23 '14 at 18:04
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/… – user148686 May 8 '14 at 12:29

$$(x/a)^2 + (y/b)^2 = 1$$
Then substitute $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$ and solve for $r(\theta)$.