1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook