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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.

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


    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

    Find the Angle of Rotation

    [tex]\cot2\theta=\frac{-1}{3}=\frac{\cot^2\theta-1}{2\cot\theta}[/tex]
    [tex]-(3\cot\theta-1)(\cot\theta+3)=0[/tex]
    [tex]\cot\theta=\frac{1}{3}[/tex]
    [tex]\theta\approx71.57 degrees[/tex]

    By drawing a triangle you can find

    [tex]\sin\theta=\frac{3}{\sqrt{10}}[/tex]
    [tex]\cos\theta=\frac{1}{\sqrt{10}}[/tex]
    Then using the 2 equations you find
    [tex]x=\frac{x'-3y'}{\sqrt{10}}[/tex]
    [tex]y=\frac{3x'+y'}{\sqrt{10}}[/tex]

    Substitute that back in and you get

    [tex]3(\frac{x'-3y'}{\sqrt{10}})^2-2\sqrt{3}(\frac{x'-3y'}{\sqrt{10}})(\frac{3x'+y'}{\sqrt{10}})+(\frac{3x'+y'}{\sqrt{10}})^2+2\frac{x'-3y'}{\sqrt{10}}+2\sqrt{3}\frac{3x'+y'}{\sqrt{10}}[/tex]

    Now I need to simplify...

    [tex]\frac{3x'^2-18x'y'+27y'^2}{10}+\frac{-6\sqrt{3}x'^2+18\sqrt{3}x'y'+6\sqrt{3}y'^2}{10}+\frac{9x'^2+6x'y'+y'^2}{10}+\frac{2x'-6y'}{\sqrt{10}}+\frac{6\sqrt{3}x'+2\sqrt{3}y'}{\sqrt{10}}[/tex]

    Simplify more...


    [tex]\frac{(12-6\sqrt{3})x'^2+(-12+18\sqrt{3})x'y'+(28+6\sqrt{3})y'^2}{10}+\frac{(2+6\sqrt{3})x'+(-6+2\sqrt{3})y'}{\sqrt{10}}[/tex]

    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

    ehild

    User Avatar
    Homework Helper

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

    ehild
     
    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.

    Because

    [tex]3\cot\theta-1=0[/tex]
    [tex]\cot\theta=\frac{1}{3}[/tex]
     
  5. Jul 18, 2010 #4

    HallsofIvy

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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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