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Math Help - rotate conic

  1. #1
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    rotate conic

    The question is how to use matrices to determine the type of curve and its characteristics. The curve is
    5x^2-4xy+2y^2-24 = 0
    Any help is appreciated
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  2. #2
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    Quote Originally Posted by delpin View Post
    The question is how to use matrices to determine the type of curve and its characteristics. The curve is
    5x^2-4xy+2y^2-24 = 0
    Any help is appreciated
    Apply a coordinate rotation to the (x, y) coordinate system:
    \left ( \begin{matrix} x \\ y \end{matrix} \right ) = \left ( \begin{matrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{matrix} \right ) \left ( \begin{matrix} x' \\ y' \end{matrix} \right )
    such that when the rotation is applied the coefficient of x'y' is zero. Then you can apply the discriminant test to the equation in x' and y' to determine the kind of curve.

    For example, in this case I get the coefficient of the x'y' term to be
    -8~cos^2(\theta) + 6~sin(\theta)~cos(\theta) + 4

    The solution here is nasty and I could only do it numerically. I got \theta \approx 0.46365~rad. The equation becomes
    6x'^2 + y'^2 - 24 = 0

    which is clearly an ellipse.

    -Dan
    Attached Thumbnails Attached Thumbnails rotate conic-ellipse.jpg  
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