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Math Help - Rotated ellipse equation sought

  1. #1
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    Smile 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!
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  2. #2
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    Is there some standard way of "rotating" a geometric figure in 2D?

    Maybe this needs to be in the geometry forum or something?
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    multiplying by this matrix will rotate the graph by the angle theta

    <br />
\begin{bmatrix}<br />
x' \\<br />
y'<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
\cos \theta && -\sin \theta \\<br />
\sin \theta && \cos \theta<br />
\end{bmatrix} <br />
\begin{bmatrix}<br />
x \\<br />
y<br />
\end{bmatrix}<br /> <br /> <br /> <br />

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

    x=3+3\cos(t) and

    y=2+2\sin(t)

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

    <br />
\begin{bmatrix}<br />
\cos \frac{\pi}{4} && -\sin \frac{\pi}{4} \\<br />
\sin \frac{\pi}{4} && \cos \frac{\pi}{4}<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
3+3\cos(t) \\<br />
2+2\sin(t) <br />
\end{bmatrix} <br /> <br /> <br />

    <br />
\begin{bmatrix}<br />
\frac{\sqrt{2}}{2} && -\frac{\sqrt{2}}{2} \\<br />
\frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}<br />
\end{bmatrix} <br />
\begin{bmatrix}<br />
3+3\cos(t) \\<br />
2+2\sin(t) <br />
\end{bmatrix}<br />

    so then

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

    I hope this helps.
    Last edited by TheEmptySet; March 14th 2008 at 06:35 PM. Reason: Need to fix lost - sign grrr!!
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  4. #4
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    Smile

    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?
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  5. #5
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    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?

    at the above link.

    I hope this is what you were also looking for .
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