MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We can rotate a point 'circularly' about an arbitrary axis:

the equation is here, but this site doesn't trust me enough yet to post an image.,

But as we walk theta 0 -> 2PI this takes the point around a "unit circle" around the axis you're rotating about

How can we make it so as theta 0 -> 2PI the results are about an ellipse of width a, height b?

I do not want to apply transformation matrices to the points after rotating them about the axis - what I'm looking for is an "elliptical" rotation matrix, if anyone knows of such a thing!

share|cite|improve this question

closed as off topic by Scott Morrison Nov 7 '09 at 17:54

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

I'm closing this as insufficiently interesting to mathematicians. Please see the FAQ, and bring any discussion over to – Scott Morrison Nov 7 '09 at 18:01
are you sure? This isn't very friendly. – bobobobo Nov 8 '09 at 19:58

Sure, you can conjugate the rotation matrix by a matrix which carries the unit circle to the ellipse in question, e.g., the diagonal 2x2 matrix with entries a and b.

share|cite|improve this answer

Sure. In 2 dimensions:

$$\begin{pmatrix}\cos\theta&k\sin\theta\\\\ -k^{-1}\sin\theta&\cos\theta\end{pmatrix}$$

The idea: Scale the $x$ axis by $k$, rotate, then scale back. Now pick $k$ appropriately (left as an exercise).

share|cite|improve this answer
Hah! Foiled by Firefox, which crashed while I was adding this answer. So Reid got there first. Oh well. 8-) – Harald Hanche-Olsen Nov 7 '09 at 16:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.