Find semi-major/semi-minor axis of ellipse from parametric equations with different phase

Question

i would like to know if it is possible to calculate the semi-major/semi-minor axis of an ellipse from these coordinates:

$$ x = \hat{x} \cdot \cos(Ωt-θ) $$

$$ y = \hat{y} \cdot \sin(Ωt-φ) $$

Answer 1

Position vector $r = (x, y) = ( \cos \Omega t - \theta, \sin \Omega t - \varphi ) $

Define the variable $t' = \Omega t - \theta $, then $\Omega t - \varphi = t' + \theta - \varphi = t' + \psi$ where $\psi = \theta - \varphi $

Hence,

$r = (x, y) = ( \cos t' , \sin(t' + \psi) ) $

Expanding the sine function,

$r = (x, y) = (\cos t' , \sin t' \cos \psi + \cos t' \sin \psi ) $

And this can be written as

$r = (x, y) = \cos t' (1, \sin \psi) + \sin t' (0, \cos \psi) $

In matrix-vector form use a $2 \times 1$ vector for $r$, i.e. $r = \begin{bmatrix} x \\ y \end{bmatrix}$, and let

$ u = \begin{bmatrix} \cos t' \\ \sin t' \end{bmatrix} $

and let

$V = \begin{bmatrix} 1 && 0 \\ \sin \psi && \cos \psi \end{bmatrix}$

Then

$r = V u$

From which $ u = V^{-1} r $

Since $u^T u= \cos^2 t' + \sin^2 t' = 1 $, then it follows that the algebraic equation of the ellipse is

$ r^T V^{-T} V^{-1} r = 1 $

Now all we have to do is find the eigenvalues of the matrix $V^{-T} V^{-1}$.

We have that

$V^{-1} = \dfrac{1}{\cos \psi} \begin{bmatrix} \cos \psi && 0 \\ -\sin \psi && 1 \end{bmatrix} $

So,

$G = V^{-T} V^{-1} = \dfrac{1}{\cos^2 \psi} \begin{bmatrix} 1 && - \sin \psi \\ -\sin \psi && 1 \end{bmatrix} $

We want to put $G$ in the form: $G = R D R^T $

Using standard diagonalization of a $2 \times 2$ matrix, and since $G_{11} = G_{22} = 1 $, then the angle between the axes of the ellipse and the coordinate axes is $\phi = \dfrac{\pi}{4} $

The diagonal matrix $D$ entries are computed as follows

$ D_{11} = G_{11} \cos^2 \phi + G_{22} \sin^2 \phi + G_{12} \sin 2 \phi $

$ D_{22} = G_{11} \sin^2 \phi + G_{22} \cos^2 \phi - G_{12} \sin 2 \phi $

These evaluate to,

$D_{11} = \dfrac{ 1 - \sin \psi }{\cos^2 \psi} = \dfrac{1}{1 + \sin \psi} $

$D_{22} = \dfrac{1 + \sin \psi }{\cos^2 \psi} = \dfrac{1}{1 - \sin \psi } $

Therefore, the semi-major axis of the ellipse has length $\sqrt{1 + \sin \psi}$ and the semi-minor axis has length $\sqrt{1 - \sin \psi} $

Answer 2

EDIT.

I'm rewriting my answer setting $\hat x=\hat y=1$, because this seems to be (?) what the OP had in mind.

The rightmost point of the ellipse is $P=(1,\sin(\theta-\phi))$, reached for $\Omega t=\theta$. Tangent at $P$ is then parallel to $y$-axis and semi-diameter $OP$ is conjugated to semidiameter $OQ$, where $Q=(0,\cos(\theta-\phi))$ is an intersection of the ellipse with $y$-axis.

Semi-axes $a$ and $b$ of the ellipse can be then computed from Apollonius' formulas: $$ \begin{align} a^2+b^2 &=OP^2+OQ^2=2\\ ab &=OP\cdot OQ\cdot \sin(\angle POQ)=\cos(\theta-\phi), \end{align} $$ which can be solved to $$ a=\sqrt{1+\sin(\theta-\phi)}, \quad b=\sqrt{1-\sin(\theta-\phi)}. $$

Answer 3

Hint:

Express the squared distance to the origin as a function of $2\Omega t$. Then find the extrema.

$$2d^2=\\(\hat x^2\cos(2\theta)-\hat y^2\cos(2\phi))\cos(2\Omega t)+(\hat x^2\sin(2\theta)-\hat y^2\sin(2\phi))\sin(2\Omega t)+\hat x^2-\hat y^2$$

$$a\cos(u)+b\sin(u)+c\in[c-\sqrt{a^2+b^2},c+\sqrt{a^2+b^2}].$$