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I couldn't figure out a way to solving this problem.

Consider the following parametric equations:

$$x = m \cos(\omega t)\qquad y = n \cos(\omega t-\phi)$$

Is this an ellipse for all real values of $m$, $n$, $\omega$, and $\phi$? If not, how can we prove that these equations do describe an ellipse barring the set of values for the above constants for which they don't?

Many thanks.

dollop15972'
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  • So, what really are we supposed to assume here and what is to be proved? It might help if you explained what you know or have been able to find out about the question, even if it did not lead to an answer. For example, do you really believe those two equations parameterize a polynomial in two variables? Why? What is the degree of the polynomial? – David K Feb 05 '22 at 16:48
  • My bad; meant to describe a general function. – dollop15972' Feb 05 '22 at 17:19
  • Please also refer to my post here for your further interest. – Ng Chung Tak Feb 06 '22 at 07:23

1 Answers1

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$ x = m \cos \omega t $

$ y = n \cos(\omega t - \phi) = n ( \cos \omega t \cos \phi + \sin \omega t \sin \phi )$

This is a linear system in $\cos \omega t $ and $\sin \omega t $, the system is given by

$ \begin{bmatrix} m && 0 \\ n \cos \phi && n \sin \phi \end{bmatrix} \begin{bmatrix} \cos \omega t \\ \sin \omega t \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} $

The solution (assuming $\sin \phi \ne 0 $ and $ n \ne 0 $ ) is

$ \begin{bmatrix} \cos \omega t \\ \sin \omega t \end{bmatrix} = \dfrac{1}{ m n \sin \phi } \begin{bmatrix} n \sin \phi && 0 \\ - n \cos \phi && m \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

Now, since $\cos^2 \omega t + \sin^2 \omega t = 1 $, then the equation governing $x, y$ is

$\begin{bmatrix} x && y \end{bmatrix} \begin{bmatrix} n^2 && - n m \cos \phi \\ - nm \cos \phi && m^2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = m^2 n^2 \sin^2 \phi $

written concisely,

$ r^T Q r = c $

The principal determinants are $n^2$ and $n^2 m^2 (1 - \cos^2 \phi) = n^2 m^2 \sin^2 \phi $

So to have a positive definite matrix $Q$ , we must also have that $m \ne 0 $

If these three conditions are met, then the above equation represents an ellipse, because then the positive definite matrix $Q$ can be diagonalized into

$Q = R D R^T $

where $D$ is the diagonal matrix of $Q$'s eigenvalues, and $R$ is the rotation matrix whose columns are the corresponding (orthogonal) unit eigenvectors. Now the equation above can be re-written as

$ r^T R D R^T r = c $

Divide through by $c$,

$ r^T R E R^T r = 1 $

with $E = \dfrac{1}{c} D $

Now let $v = R^T r $ so that $ r = R v $ , then

$ v^T E v = 1 $

and this is clearly an equation of an ellipse because the diagonal entries of $E$ are both positive. Finally the vector $r$ is just a rotation about the origin of vector $v$, thus it describes a rotated ellipse whose angle of rotation is determined by the angle between the first column of $R$ and the positive $x$ axis.

Hosam Hajeer
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