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Rotation of an ellipse

  1. Jan 13, 2016 #1
    Hello! Okay- This is a relatively simple problem, but for some reason I'm having huge difficulty with it.

    So I have the equation of an ellipse, x^2-6sqrt3 * xy + 7y^2 =16, which I have converted into quadratic form to get (13, -3sqrt3, -sqrt3, 7) and I need to rotate it using the normal rotation matrix in two dimensions (cos, -sin, cos, sin)

    But Im struggling to actually apply the rotation matrix- Do apply it to the quadratic form of the matrix? And if so how can I extract the equation of the rotated ellipse from this?

    Any help would be GREATLY appreciated!! Thank you
     
  2. jcsd
  3. Jan 13, 2016 #2

    blue_leaf77

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    Assume the equation of ellipse you have there to be written in the already rotated coordinate system ##(x',y')##, thus
    $$
    x'^2-6\sqrt{3} x'y' + 7y'^2 =16
    $$
    To obtain the expression of this same ellipse in the unrotated coordinate system, you have to apply the clockwise rotation matrix to the point ##(x',y')##. By the way the correct rotation matrix is
    $$
    \left( \begin{array}{cc}
    \cos \theta & \sin\theta \\
    -\sin\theta & \cos \theta \\
    \end{array} \right)
    $$
    where ##\theta## is positive for counterclockwise rotation.
     
    Last edited: Jan 14, 2016
  4. Jan 13, 2016 #3

    Okay, so do I simply apply the rotation matrix to the quadratic form?
     
  5. Jan 13, 2016 #4

    blue_leaf77

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    First of all, how did you get that quadratic form, that doesn't seem to lead to the original ellipse equation.
     
  6. Jan 13, 2016 #5
    if you multiple by (x,y)T A (x,y) where A is the matrix stated above. Sorry I don't know how to use matrix notation on here- but the T represents transpose
     
  7. Jan 14, 2016 #6

    blue_leaf77

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    Well yeah I know that, but if you really carry out the matrix operation
    $$
    \left( \begin{array}{cc}
    x & y \\
    \end{array} \right)
    %
    \left( \begin{array}{cc}
    13 & -3\sqrt{3} \\
    -\sqrt{3} &7 \\
    \end{array} \right)
    %
    \left( \begin{array}{c}
    x \\
    y
    \end{array} \right) =
    \left( \begin{array}{cc}
    x & y \\
    \end{array} \right)
    %
    \left( \begin{array}{c}
    13x-3\sqrt{3}y \\
    -\sqrt{3}x + 7y \\
    \end{array} \right) =
    13x^2 - 4\sqrt{3}xy + 7y^2
    $$
    It cannot be the same as the ellipse equation you have in your first post for whatever constant value in the right hand side.
     
  8. Jan 14, 2016 #7
    Sorry- that was a typo on my part- both of the non-leading diagonal entries should be -3sqrt3
     
  9. Jan 14, 2016 #8

    blue_leaf77

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    What about the coefficient of ##x^2##?
     
  10. Jan 14, 2016 #9
    Also a typo- sorry
    it is meant to be 13x^2
     
  11. Jan 14, 2016 #10
    I unfortunately don't know how to edit my original post to correct these though
     
  12. Jan 14, 2016 #11

    blue_leaf77

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    What are you asked to do? Are you asked to express the rotated ellipse in quadratic form or in algebraic form?
     
  13. Jan 14, 2016 #12
    algebraic, but in the unrotated co-ordinate system
     
  14. Jan 14, 2016 #13

    blue_leaf77

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    Then you don't need to take a detour by going to the quadratic form since you already have the algebraic form of the original ellipse. So as I said earlier, imagine you have the ellipse in the already rotated coordinate system ##(x',y')##, which means
    $$
    13x'^2-6\sqrt{3} x'y' + 7y'^2 =16
    $$
    At the same time, you also have the matrix relation between ##(x',y')^T## and ##(x,y)^T##, related by coordinate rotation. From this, you should be able to obtain expressions for ##x'## and ##y'## in terms of ##x## and ##y## and substitute to the ellipse equation.
     
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