1

If a non-axis-aligned ellipsoid is defined by $$\tag{1} \bf x^T A x=1$$ where $\bf x$ $=[x, y, z]^T$, $\bf A$ $=\begin{bmatrix}a_{xx}&a_{xy}&a_{xz}\\a_{yx}&a_{yy}&a_{yz}\\a_{zx}&a_{zy}&a_{zz} \end{bmatrix}$, and $a_{ij}=a_{ji}$,

then Eqn(1) is the following when expanded: $$\tag{2} a_{xx}x^2+a_{yy}y^2+a_{zz}z^2+2a_{xy}xy+2a_{xz}xz+2a_{yz}yz=1$$

at $z=0$ the coordinates of the ellipsoid is: $$\tag{3} a_{xx}x^2+a_{yy}y^2+2a_{xy}xy=1$$

which I think is also an equation of an ellipse since its the intersection of the ellipsoid and the x-y-plane.

Equation 3 is in implicit form. How can its parametric equations be found?

Ted Shifrin
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Armadillo
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  • @EthanBolker possibly, however in that question the ellipsoid is axis-aligned. – Armadillo Apr 29 '19 at 20:41
  • In the possible duplicate the ellipsoid is axis aligned but the plane is arbitrary. If you rotate to align your ellipsoid (straightforward linear algebra) the $s-y$-plane will rotate and be unaligned and can use that solution. – Ethan Bolker Apr 29 '19 at 20:45
  • See also https://math.stackexchange.com/q/596016/265466 for how to convert the equation of the ellipse to canonical form, from which a standard parameterization can be extracted. The ellipsoid and its orientation are irrelevant to your problem at this point. Or, just find a pair of conjugate semi-diameters $\mathbf u$ and $\mathbf v$. A parameterization of the ellipse is then just $\vec u\cos t+\vec v\sin t$. – amd Apr 29 '19 at 21:02
  • The right tag is linear algebra, not differential geometry. – Ted Shifrin Apr 29 '19 at 22:31

2 Answers2

1

Diagonalize the top-left $2\times2$ matrix, and you will reduce to a form

$$\lambda u^2+\mu v^2=1$$ where $\lambda,\mu$ are the two Eigenvalues, which are positive. Then

$$u=\frac{\cos\theta}{\sqrt\lambda},v=\frac{\sin\theta}{\sqrt{\mu}}$$ and $x,y$ are the linear combinations of $u,v$ given by the Eigenvectors.

0

It suffices to transform in a first step

$$a_{xx}x^2+a_{yy}y^2+2a_{xy}xy=1\tag{0}$$

into a polar equation using ... polar coordinates

$$x=r \cos \theta, \ y=r \sin \theta, \tag{1}$$

giving :

$$r^2(a_{xx}\cos^2 \theta +a_{yy}\sin^2 \theta+a_{xy}(2 \sin \theta \cos \theta))=1.\tag{2}$$

Using formulas

$$\begin{cases}\sin 2 \theta &=& 2 \sin \theta \cos \theta \\ \cos^2 \theta&=&\tfrac12(1+\cos 2 \theta) \\ \sin^2 \theta&=&\tfrac12(1-\cos 2 \theta)\end{cases}$$

one obtains from (2) the following polar equation :

$$r=\dfrac{1}{\sqrt{A \cos(2 \theta)+B+a_{xy} \sin (2 \theta)}}\tag{3}$$

with $A:=\tfrac12(a_{xx}-a_{yy})$ and $B:=\tfrac12(a_{xx}+a_{yy})$.

(See remark below).

Having polar equation (3), it is now straightforward to convert it into a parametric equation by using again formulas (1) but with the expression of $r$ given by (3).

Remark : instead of (3), we could have taken as well

$$r=\dfrac{-1}{\sqrt{...}}\tag{4}.$$

Indeed, equation (0) is homogeneous with degree 2, meaning that it is invariant when $(x,y)$ is replaced by its opposite $(-x,-y)$ ; as a consequence, (4) generates the very same curve.

Jean Marie
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