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Rotating a Parabola

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Rotate the axis to eliminate the xy-term.


    2. Relevant equations

    3. The attempt at a solution

    Find the Angle of Rotation

    [tex]\theta\approx71.57 degrees[/tex]

    By drawing a triangle you can find

    Then using the 2 equations you find

    Substitute that back in and you get


    Now I need to simplify...


    Simplify more...


    Phew.. that was a lot of typing... Now I could cross multiply and type all of that out as well but if I did my x'y' terms won't cancel out and since that was the whole purpose of this endeavor I went wrong somewhere. Do you know where I made my mistake?
  2. jcsd
  3. Jul 18, 2010 #2


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    Homework Helper

    Your angle is wrong. How did you get it? Did not you miss a square root?

    Last edited: Jul 18, 2010
  4. Jul 18, 2010 #3
    I saw how I was wrong before and used the angle [tex]\frac{1}{\sqrt{3}}[/tex] and it worked out fine. However, looking back at it again I cant find why I took the square root.


  5. Jul 18, 2010 #4


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    Science Advisor

    ehild is right.

    The formula you start with (although the LaTex doesn't appear to be right) is
    [tex]cot(2\theta)= \frac{A- C}{B}[/tex]

    And here, A= 3, C= 1, ande [itex]B= 2\sqrt{3}[/itex]

    [tex]cot(2\theta)= \frac{3- 1}{2\sqrt{3}}= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}[/tex]
    not "1/3".

    You didn't take the square root- it is already in the equation: [itex]2x^2+ 2\sqrt{3}xy+ y^2[/itex]
  6. Jul 18, 2010 #5
    Ok, Now I see where I went wrong. Thanks.
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