I have found here that an ellipse in the 3D space can be expressed parametrically by

$$\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$$

with $\mathbf c = (c_1,c_2,c_3)$ being the center of the ellipse and the lenghts of the half-axis being the lengths of the vectors $\mathbf u = (u_1,u_2,u_3)$ and $\mathbf v = (v_1,v_2,v_3)$.

How could these three vectors $\mathbf c$, $\mathbf u$ and $\mathbf v$ be related to the directions of the axis of the ellipse? Is there maybe a parametric equation for the ellipse in an arbitrary plane of the space whose elements have a more intuitive meaning?

  • $\begingroup$ Hint : Points on ellipse can be parameterized as $(a\cos t, b\sin t)$. $\endgroup$
    – cosmo5
    Jan 21, 2021 at 20:49
  • $\begingroup$ You have two vectors which represent the direction of both major and minor semi-axes and its magnitude. You have a position vector which represents its center and both vectors are orthogonal to the normal vector of the plane that ellipse lies in. This represents all points on the ellipse. $\endgroup$
    – Math Lover
    Jan 21, 2021 at 20:58

1 Answer 1


In the parametric equation $\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$, we have: $\mathbf c$ is the center of the ellipse, $\mathbf u$ is the vector from the center of the ellipse to a point on the ellipse with maximum curvature, and $\mathbf v$ is the vector from the center of the ellipse to a point with minimum curvature. I assumed $\|u\| > \|v\|$.

  • 3
    $\begingroup$ If you don't assume that $u \perp v$ you will not have in general maximal/minimal curvature at the extremities of $u$ and $v$. $\endgroup$
    – Jean Marie
    Jan 21, 2021 at 21:53
  • 2
    $\begingroup$ You are correct. If they are not perpendicular, they are just some vectors from the center to the curve. But for any given ellipse you can choose $c,u,v$ as I specified to get a parameterization. $\endgroup$
    – GEdgar
    Jan 21, 2021 at 22:50
  • 1
    $\begingroup$ I plotted this parametrization along with the $\mathbf u$ and $\mathbf v$ vectors: it's true that they are not in the direction of the axis of the ellipse unless they are orthogonal (although they are indeed in the plane of the ellipse). math3d.org/FMjy6jzb $\endgroup$
    – Invenietis
    Jan 22, 2021 at 9:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .