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: Ellipse question

  1. Sep 10, 2004 #1
    Ack, such a simple question, but I haven't worked with conic sections in years. Can anyone suggest an elegant way to show that

    [tex]x=f*Sin(wt+\theta)[/tex]
    [tex]y=g*Sin(wt+\phi)[/tex]

    is an ellipse? I've tried using a rotation matrix on standard parametric ellipse equations and then solving for the angle of rotation and the axes sizes in terms of the variables but it seems messy. Then I tried getting it to fit the general equation but I'm not sure how that would work. However, a simpler method is eluding me. Any help?
     
  2. jcsd
  3. Sep 10, 2004 #2

    krab

    User Avatar
    Science Advisor

    Some hints: This is not an upright ellipse like [itex]x^2/a^2+z^2/b^2=1[/itex], unless [itex]\phi-\theta=\pi/2[/itex]. But you can get it into that form by applying a transformation like z=y+kx, where k is a constant. By expanding the sines, find k in terms f, g, [itex]\phi[/itex] and [itex]\theta[/itex] (actually will depend only on [itex]\phi-\theta[/itex] and f/g), and you are there.
     
  4. Sep 11, 2004 #3
    Can you clarify what you mean by a transformation of the form z=y+kx? what is y?
     
  5. Sep 11, 2004 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    If you don't mind a little nonrigorous math you can try the following.

    Rewriting your equations slightly:

    [tex]x = A cos(\omega t + \alpha)[/tex]
    [tex]y = B cos(\omega t + \beta)[/tex]

    The second equation gives
    [tex]\omega t = \cos ^{-1} \left(\frac {y}{B}\right) - \beta [/tex]

    Substitute into the first equation:
    [tex]x = A \cos( \cos ^{-1} \frac {y}{B} + \alpha - \beta)[/tex]

    Use the addition formulas for cosine and the basic identities for the inverse trig function and the equation can be cast into the familiar elliptical form.
     
  6. Sep 11, 2004 #5
    By familiar elliptical form, do you mean general form? I can't seem to get rid of that square root term from the inverse trig identity...
     
  7. Sep 11, 2004 #6

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Rearrange terms to get the radical on one side of the equation then square both sides. The "general form" is [itex]ax^2+bxy+cy^2+d=0[/itex] and, depending on the signs and values of the various coefficients, will produce an ellipse, hyperbola or parabola.
     
  8. Sep 11, 2004 #7
    I got it, but thanks for your replies.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook