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Homework Help: Second Degree Equations, xy nonzero, rotation of axes question

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] x^{2}-4xy+y^{2} - 1= 0 [/tex]

    2. Relevant equations

    Rotation of axes equations:

    [tex] x = X cos(\alpha)-Ysin(\alpha) [/tex]

    [tex] y = Xsin(\alpha) + Ycos(\alpha) [/tex]

    Alpha is given in the text as:

    [tex] cot(2\alpha) = \frac {A-C}{B} [/tex]

    but I used the fact that:

    [tex] cot(\alpha) = \frac {1}{tan(\alpha)} [/tex]

    to solve for alpha like so:

    [tex] \alpha = \frac {tan^{-1}(\frac{B}{A-C})}{2} [/tex]

    3. The attempt at a solution

    ok so i try to find alpha first like:

    [tex] \alpha = \frac {tan^{-1}(-\frac{4}{1-1})} {2} [/tex]

    which means that it is really just:

    [tex] \alpha = \frac {tan^{-1}(0)} {2} [/tex]

    and the inverse tangent of 0 is 0 degrees, the only problem is that I am looking for 45 degrees, where did I go wrong? Is there something wrong with me using the trig relationship to solve for alpha?



    I just used the same alpha equation with an inverse tangent on the equation:

    [tex] x^{2}+2\sqrt{3}xy-y^{2}-7=0 [/tex]

    and got alpha at 30 degrees and a simplified equation of:

    [tex] 2x^{2}-2y^{2}-7=0 [/tex]

    which all passes discrimination tests so what exactly is wrong with the first problem? of course this only worked because A-C was nonzero...
    Last edited: Jul 19, 2010
  2. jcsd
  3. Jul 19, 2010 #2


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    Homework Helper

    It's not inverse tangent of 0. It's inverse tangent of -4/0. -4/0 is undefined.

    I probably would have just worked with the cotangent equation:
    cot(2\alpha) &= \frac {A-C}{B} \\
    &= \frac{1 - 1}{-4} \\
    &= 0

    If 0 < α < π/2, then 0 < 2α < π, and the only place between 0 and π where the cotangent equals 0 is at π/2. So
    2\alpha &= \frac {\pi}{2} \\
    \alpha &= \frac{\pi}{4}
    ... which is the angle you wanted.

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