# Equation of ellipsoid

1. Sep 27, 2009

### 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.

2. Sep 27, 2009

### 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 $x,y$.

3. Sep 28, 2009

### 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 :)