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I'm creating a computer program where I need to calculate the parametrized circumference of an ellipse, like this:

x = r1 * cos(t), y = r2 * sin(t) 

Now, say I want this parametrized ellipse to be tilted at an arbitrary angle. How do I go about this? Any obvious simplifications for i.e 30, 45 or 60 degrees?

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If you want to rotate $\theta$ radians, you should use $t\mapsto \left( \begin{array}{c} a \cos (t) \cos (\theta )-b \sin (t) \sin (\theta ) \\ a \cos (t) \sin (\theta )+b \sin (t) \cos (\theta ) \end{array} \right) $

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If you rotate $(x=r_1\cos t,y=r_2\sin t)$ by $\theta$ about $(0,0)$, the resulting curve is given by (x'=r_1\cos t\cos\theta-r_2\sin t\sin\theta, y'=r_1\cos t\sin\theta+r_2\sin t\cos\theta).

(Using the fact that complex multiplication by $e^{i\theta}$ rotates by $\theta$ about $0$, the point $(x,y)=x+yi$ is mapped to $(x+iy)(\cos\theta+i\sin\theta)=$ $x\cos\theta-y\sin\theta+i(x\sin\theta+y\cos\theta)=$ $(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$.)

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    Thanks! I'll try this out in code tomorrow.2011-01-14
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Update:

Given

$x = r_1 \cos(t)$, $y = r_2 \sin(t)$

using the equation of ellipse in polar form we have : (could someone confirm this please?)

$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta)}+{r_2^2 \cos^2 (\theta)}}} $

to rotate by angle $\phi$ :

$ r(\theta) = \frac {r_1r_2}{\sqrt {{r_1^2 \sin^2 (\theta-\phi)}+{r_2^2 \cos^2 (\theta-\phi)}}} $

The original incorrect answer is below:

$x = r_1 \cos(t + \phi)$, $y = r_2 \sin(t + \phi) \text {, where }\phi$ is your angle of tilt in radians

This was as pointed out was my mistake on thinking polar while working with parametric. I have tried converting this to polar form but adding angle of $\phi$ and reverting back to parametric form so far has not been possible.

Has anyone seen solution of this problem by converting to polar coordinates and performing an angle addition and converting back to parametric form?

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    @Isaac : Thanks, fixed!, Also turned it into community wiki ( is that good ?)2011-01-24