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Equation of ellipsoid

  1. Sep 27, 2009 #1
    given an ellipsoid in parametric form in t, I was trying to get to the classical equation in x,y. Things are very straightforward, as long as the ellipse radii are aligned with the principal axes. Instead, I am trying to find the equation of a "rotated" ellipse, given a parametrization in t.

    I tried the following... Let's define the position vector:

    [tex]\mathbf{r}(t) = \mathbf{a}cos(t) + \mathbf{b}sin(t)[/tex]

    [tex]\mathbf{a}=a_1\mathbf{e_1} + a_2\mathbf{e_2}[/tex]
    [tex]\mathbf{b}=b_1\mathbf{e_1} + b_2\mathbf{e_2}[/tex]

    and we have that [tex]<\mathbf{a},\mathbf{b}>=0[/tex], that is, the directional radii are perpendicular but not aligned to the main axes.
    Since [tex]x = <\mathbf{r},\mathbf{e_1}>[/tex], and [tex]y = <\mathbf{r},\mathbf{e_2}>[/tex], we have:

    [tex]x = a_1cos(t) + b_1sin(t)[/tex]
    [tex]y = a_2cos(t) + b_2sin(t)[/tex]

    At this point I got stuck, because I can't manage to get rid of t. When the ellipse is aligned to the main axes we have [tex]b_1=0[/tex], and [tex]a_2=0[/tex], and everything becomes easy by squaring the terms.
    I know that the final result should be of the form: [tex]\mathbf{x^T}A\mathbf{x}[/tex] where A is symmetric positive definite, but I can't really get there.
  2. jcsd
  3. Sep 27, 2009 #2
    Your ellipse is centered at the origin.
    You have two equations (linear in [itex] \cos t[/itex] and [itex] \sin t[/itex]). Solve them like this: [itex] \cos t = ??[/itex], [itex] \sin t = ??[/itex], both right-hand-sides linear in [itex]x,y[/itex]. Then take the equation [itex]\sin^2 t + \cos^2 t = 1 [/itex], substitute in your results, you get something quadratic in [itex] x,y[/itex].
  4. Sep 28, 2009 #3
    Thanks a lot!
    I can't believe I didn't immediately find such an easy solution! It has been under my eyes all the time (even on my notes) but yesterday I simply missed it :/ ... I should punish myself now :)
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