Ellipsoid Equation: Finding the Classical Form with Rotated Radii

[mnb96]

Hi,
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:

\mathbf{r}(t) = \mathbf{a}cos(t) + \mathbf{b}sin(t)

where:
\mathbf{a}=a_1\mathbf{e_1} + a_2\mathbf{e_2}
\mathbf{b}=b_1\mathbf{e_1} + b_2\mathbf{e_2}

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

x = a_1cos(t) + b_1sin(t)
y = a_2cos(t) + b_2sin(t)

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 b_1=0, and a_2=0, and everything becomes easy by squaring the terms.
I know that the final result should be of the form: \mathbf{x^T}A\mathbf{x} where A is symmetric positive definite, but I can't really get there.

 

[g_edgar]

Your ellipse is centered at the origin.
You have two equations (linear in \cos t and \sin t). Solve them like this: \cos t = ??, \sin t = ??, both right-hand-sides linear in x,y. Then take the equation \sin^2 t + \cos^2 t = 1, substitute in your results, you get something quadratic in x,y.

 

[mnb96]

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 :)