Rotated ellipse equation sought

• Mar 14th 2008, 01:05 PM
Dirlewanger
Rotated ellipse equation sought
I'm looking for a Cartesian equation for a rotated ellipse.

h is x-koordinate of the center of the ellipse.
k is y-koordinate of the center of the ellipse.
a is the ellipse axis which is parallell to the x-axis when rotation is zero.
b is the ellipse axis which is parallell to the y-axis when rotation is zero.
phi is the rotation angle.

Here is a cartesian equation for a non-rotated ellipse:
(How do I "put rotation phi" into this?)

Code:

(x-h)^2  (y-k)^2 ------- + ------- = 1   a^2      b^2

Here is a parametric form of a rotated ellipse:

x = h+a*cos(t)*cos(phi)-b*sin(t)*sin(phi)
y = k+b*sin(t)*cos(phi)+a*cos(t)*sin(phi)

I'd like to have the x and the y in the same equation, and without the parameter t. I think that this means that I want a Cartesian equation for a rotated ellipse, but I'm sorry if I have misunderstood Cartesian here...

Thanks beforehand and a jolly good weekend to ya' all!
Cheers!
• Mar 14th 2008, 05:20 PM
Dirlewanger
Is there some standard way of "rotating" a geometric figure in 2D?

Maybe this needs to be in the geometry forum or something?
• Mar 14th 2008, 06:02 PM
TheEmptySet
multiplying by this matrix will rotate the graph by the angle theta

$\displaystyle \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

where x' and y' are the new rotated coordinates
so now suppose that we have the equations of an ellispe

$\displaystyle x=3+3\cos(t)$ and

$\displaystyle y=2+2\sin(t)$

so if we want to rotate the eqations by 45 degrees or $\displaystyle \frac{\pi}{4} radians$ we get..

$\displaystyle \begin{bmatrix} \cos \frac{\pi}{4} && -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} && \cos \frac{\pi}{4} \end{bmatrix} \begin{bmatrix} 3+3\cos(t) \\ 2+2\sin(t) \end{bmatrix}$

$\displaystyle \begin{bmatrix} \frac{\sqrt{2}}{2} && -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} 3+3\cos(t) \\ 2+2\sin(t) \end{bmatrix}$

so then

$\displaystyle x'=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(t)-\sqrt{2}-\sqrt{2}\sin(t)$
and
$\displaystyle y'=\sqrt{2}+\sqrt{2}\sin(t)+ \frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(t)$

I hope this helps.
• Mar 14th 2008, 07:43 PM
Dirlewanger
Actually, you demonstrated how to derive the parametric equations of a rotated ellipse, which I posted, from the parametric equations of a non-rotated ellipse. I can follow the matrix multiplication, yes, but is it possible to put that together in the way I'm looking for:

Joining x and y in one single equation, and getting rid of that t parameter? If I try to substitute and solve from the parametric equation system, I end up with a mess. There must be a much simpler way.

For a non-rotated ellipse, it looks quite simple:

Code:

(x-h)^2  (y-k)^2 ------- + ------- = 1   a^2      b^2

How do I turn THAT phi degrees?
(Worried)
• Mar 16th 2008, 05:14 PM
schnell
Hi,
i also had the same problem. I found the solution at the following link. I do not know whether the external links are allowed on this forum or not...

The Quadratic Curves, Circles and Ellipses FAQ

Look for the question ---

Q21. How do I calculate the coefficients of an arbitary rotated ellipse?