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I need to parametrize an ellipse centered at the origian in cartesian coordinates given by

$a x^2 + b xy + c y^2 = 1$

Now I dug up from old notes that this can be brought into normal form through a rotation $\tan 2 \theta = \frac {b}{a-c}$, which eliminates the $xy$ term in rotated coordinates $x' = x \cos \theta + y \sin \theta, y = - x \sin \theta + y \cos \theta$ and we get

$\left( \frac{a + c + \sqrt{(a-c)^2+b^2}}{2}\right) x'^2 + \left( \frac{a + c - \sqrt{(a-c)^2+b^2}}{2}\right)y'^2 = 1$

which can be expressed as a normal equation $\frac{x'^2}{\alpha^2} + \frac{y'^2}{\beta^2} = 1$ if $4ac > b^2$ and both of the coefficients above have positive signs (which we may assume). This is then parametrized in the standard way by

$x' = \alpha \cos t, y' = \beta \sin t$ for $0\le t < 2\pi$

in the rotated coordinates and by $x = x' \cos \theta - y' sin \theta, y = x' sin \theta + y' \cos \theta$ in the original coordinates. Now while the above works, it's been pointed out to me that I am making things more complicated than necessary. Are there better ways of achieving this? I was thinking of diagonalizing the matrix

$(x, y)' \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} (x,y)=1$

but I dont think that would make things easier. Is there a direct parametrization of the ellipse in terms of $a,b,c$?

Chris_77
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  • Not sure what sort of a parameterization you’re looking for, but you can parameterize an origin-centered ellipse as $\mathbf u\cos t+\mathbf v\sin t$, where the vectors $\mathbf u$ and $\mathbf v$ are conjugate radii. – amd Nov 12 '18 at 05:48
  • See another answer here. – Ng Chung Tak Nov 12 '18 at 08:35

1 Answers1

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Rewriting equation is quicker : $4a^2x^2 + 4abxy + 4acy^2 = 4a$ (multiplying by $4a$) ; $(2ax + by)^2 - b^2y^2 + 4acy^2 = 4a$ (it is a trick removing the $xy$ term) ; $(2ax + by)^2 + (4ac - b^2)y^2 = 4a$, then you can compare with equation $X^2/p + Y^2/q = 1$ of a conic (ellipse or hyperbola) for which you know a parametrization.

Majid
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Dominique Laurain
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