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Math Help - Conic equations and rotations

  1. #1
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    Conic equations and rotations

    Help with these two problems would be great:

    Find the angle of revolution in degrees needed to put the conic in standard form. 2x^2-xy-3y^2-2x+4y-6=0
    My work:

    Modeling Cot2A= (A-C)/B
    Cot2A= (2-(-)3)/-1 so Cot2A= (2+3)/-1 = 5/-1 = -5
    I am very confused on how to go from here get the angle in degrees from a -5, any help please?
    I know this isn't right but here goes,
    Cot2A=-5 so Cot A = -25 degrees???


    Find the equation for the conic in standard form and identify the conic.
    x= 5 + 2 Cos t, y= -2 + sin t, 0 ≤ t ≤ 2π.

    Do not really know what this means or where to start...
    Thanks a lot for help!
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  2. #2
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    Quote Originally Posted by Sxon View Post
    Help ... would be great:

    ...

    Find the equation for the conic in standard form and identify the conic.
    x= 5 + 2 Cos t, y= -2 + sin t, 0 ≤ t ≤ 2π.

    Do not really know what this means or where to start...
    Thanks a lot for help!
    Eliminate the parameter t in the system of equations:

    \left|\begin{array}{l}x=5+2\cos(t)\\y=-2+\sin(t)\end{array}\right. ~\implies~ \left|\begin{array}{l}\dfrac12(x-5)=\cos(t)\\y+2=\sin(t)\end{array}\right.

    Now square both equations and add: (Keep in mind that \sin^2(t) + \cos^2(t) = 1 )

    \left|\begin{array}{l}\dfrac14(x-5)^2=\cos^2(t)\\(y+2)^2=\sin^2(t)\end{array}\right  . ~\implies~
    \dfrac{(x-5)^2}4 + \dfrac{(y+2)^2}1=1

    which is the equation of an allipse.
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