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We can write any point on the circle as $(r\cos\theta,r\sin\theta$), Can we do samething for the ellipse?

enter image description here

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How do you usually parametrize (i.e. write) an ellipse? –  Git Gud May 16 '13 at 6:15
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Hint: Ellipse is nothing but a circle on axes where the units are not the same. Plug in $x=av$ and $y=bu$ in $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, you get an equation $u^2+v^2=1$. –  Inceptio May 16 '13 at 6:16
    
There are various ways to do it, depending where you want the polar origin to be, at the centre as in your picture, or at a focus, or elsewhere. –  André Nicolas May 16 '13 at 6:16
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the picture is very confusing. because $\theta$ is not an axis in $xy$-plane and elipse this way is just in some direction deformed circle. –  V-X May 16 '13 at 6:17

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The parametrization of ellipse:

$$ \begin{cases} x=a\cos\theta, \\ y=b\sin\theta. \end{cases} $$

For the circle $a=b=r$.

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A point on the ellipse is typically represented as $(a \cos(\theta), b \sin(\theta))$, where $a$ is the length of the semi-major (typically $X$) axis and $b$ is the length of the semi-minor (typically $Y$) axis. If you want to write the equation in polar form, then we have $$r = a \sqrt{1-e^2 \sin^2(\theta)}$$ where $e$ is the eccentricity defined as $e = \sqrt{1-\left(\dfrac{b}a \right)^2}$, where $b \leq a$.

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With $a$ the major and $b$ the minor semiaxis, you get $(a\cos\theta,b\sin\theta)$ for your orientation. A rotated ellipse will be a bit harder, for this I'd compute the point in its unoriented form and then rotate afterwards. Note that the angle $\theta$ in the above is a parameter, but is not actually the angle as it is in the circular case.

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A rotation just changes the angle by a constant amount. Here $\theta$ is a co-ordinate not merely a parameter - and is the angle relative to a fixed direction. –  Mark Bennet May 16 '13 at 6:19
    
@MarkBennet: Since you won't get $\tan\theta=\frac{a\sin\theta}{b\cos\theta}$ in general, it is not the angle in the usual sense. And offsetting $\theta$ won't rotate the ellipse as long as you keep $a$ associated with the $x$ value only and $b$ associated with $y$. –  MvG May 16 '13 at 6:29
    
Sorry, I had misunderstood your point - you are right. –  Mark Bennet May 16 '13 at 6:36

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