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Rotating a Hyperbola

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Rotate the axis to eliminate the xy-term. Sketch the graph of the equation showing both sets of axis.

    xy-2y-4x=0


    2. Relevant equations
    [tex]\cot2\theta=\frac{A-C}{B}[/tex]
    [tex]x=x'\cos\theta-y'\sin\theta[/tex]
    [tex]y=x'\sin\theta+y'\cos\theta[/tex]

    3. The attempt at a solution

    xy-2y-4x=0

    First I find the angle of the x' y' axis.
    using
    [tex]\cot2\theta=\frac{A-C}{B}[/tex]

    I find it to be[tex]\theta=\frac{\pi}{4}[/tex]

    Then find the x' and y' components

    by using the second 2 equations I listed I come out with.
    [tex]x=\frac{x'-y'}{\sqrt{2}}[/tex]
    [tex]y=\frac{x'+y'}{\sqrt{2}}[/tex]


    Then comes the substitutions and simplifying.

    [tex](\frac{x'-y'}{\sqrt{2}})(\frac{x'+y'}{\sqrt{2}})-2(\frac{x'+y'}{\sqrt{2}})-4(\frac{x'-y'}{\sqrt{2}})=0[/tex]

    [tex]\frac{x'^2+y'^2}{2}+\frac{-6x'+2y'}{\sqrt{2}}=0[/tex]

    [tex]\sqrt{2}(x')^2+\sqrt{2}(y')^2-12x'+4y'=0[/tex]

    I complete the square and end up with

    [tex](x'-\frac{12}{\sqrt{2}})^2+(y'+\frac{2}{\sqrt{2}})^2=20\sqrt{2}[/tex]

    then I would divide though to get a 1 on the RHS but this is wrong.

    The answer should be.

    [tex] \frac{(x'-3\sqrt{2})^2}{16}-\frac{(y'-\sqrt{2})^2}{16}=1[/tex]

    where did I go wrong?
     
  2. jcsd
  3. Jul 17, 2010 #2
    Never mind. I found my problem.
     
  4. Jul 17, 2010 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct up to here, but you mixed + and - after, and you made a mistake when multiplying the equation with 2.


    ehild
     
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