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Math Help - Rotation of a conic?

  1. #1
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    Rotation of a conic?

    I am having some difficulty answer this type of question.

    x^2-5x+y^2=3 rotated by pi/3

    cos pi/3= 1/2 sin pi/3= sqrt{3}/2

    But I am unsure how to proceed my book says

    x with x cos theta + y sin theta

    y with -x sin theta + y cos theta

    Then plug this in the original equation

    I think the graph is a circle
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The transformed point of (x,y) by means of the rotation is

    \begin{Bmatrix} x'=\dfrac{1}{2}x-\dfrac{\sqrt{3}}{2}y\\{}\\y'=\dfrac{\sqrt{3}}{2}x+  \dfrac{1}{2}y\end{matrix}

    Equivalently

    \begin{Bmatrix} x=\dfrac{1}{2}x'+\dfrac{\sqrt{3}}{2}y'\\y=-\dfrac{\sqrt{3}}{2}x'+\dfrac{1}{2}y'\end{matrix}

    Now, substitute in the original conic.
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