Angle of ellipse's long axis and abscisa

Question

I am reading a paper where I encountered the following equations

$x=a.\cos(\omega t)$ and $y=b.\cos(\omega t + \phi)$
$\phi$ and $\omega$ are constants, t is the variable parameter.

Then the paper suggests this is the parametric equation of an ellipse.

First of all, how can I prove this?
Then the area of the ellipse is calculated and is found as $A= \pi a b \sin \phi$, which I found by using Green's theorem. Is there a more elegant way to find this?

Then, last but not least, the angle $\gamma$ between the long axis and the abscisa needs to be determined and according to the paper it's given by
$\tan 2\gamma = \frac{2ab \cos \phi}{a^2 - b^2}$.

Can somebody prove this please?

Answer 1

I think that your proof of the area is calculated most elegantly with Green's Theorem since you have a parametric form.

For the angle, I think the fastest way depends on how much you know already about the ellipse, for example do you know that the center is the origin. To avoid making assumptions without verifying them, I would look for the rotation transforms the standard parametric form $(p \cos u,q \sin u)$ into your parametric form: $$ \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} \cos\gamma & -\sin\gamma\\ \sin\gamma & \cos\gamma \end{pmatrix} \begin{pmatrix} p \cos u\\ q \sin u \end{pmatrix} = \begin{pmatrix} p \cos u\cos\gamma-q \sin \gamma \sin(u)\\ p \cos u\sin\gamma+q \cos \gamma \sin(u) \end{pmatrix} $$ $$ = \begin{pmatrix} p \cos \gamma \left(\cos u-\frac{q \sin \gamma}{p \cos \gamma} \sin u \right)\\ p \sin \gamma \left(\cos u+\frac{q \cos \gamma}{p \sin \gamma} \sin u \right) \end{pmatrix} $$

Now say $$ \tan\theta_1=\frac{q}{p} \tan \gamma, $$ $$ \tan\theta_2=-\frac{q}{p} \cot \gamma. $$ Then you can verify that $$ \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} p \frac{\cos \gamma} {\cos \theta_1} \cos(u+\theta_1)\\ p \frac{\sin \gamma} {\cos \theta_2} \cos(u+\theta_2) \end{pmatrix} $$ Now subsitute $u=\omega t -\theta_1$: $$ \begin{pmatrix} p \frac{\cos \gamma}{ \cos \theta_1} \cos(\omega t)\\ p \frac{\sin \gamma}{ \cos \theta_2} \cos(\omega t-\theta_1+\theta_2) \end{pmatrix} $$ We identify that $\phi=\theta_2-\theta_1$, $a=p \frac{\cos \gamma}{ \cos \theta_1}$,$b=p \frac{\sin \gamma}{ \cos \theta_2}$. Taking cos of both sides of the first: $$ \cos\phi=\cos(\theta_2- \theta_1)=\cos \theta_2 \cos \theta_1+\sin \theta_2 \sin \theta_1=\cos \theta_2 \cos \theta_1 (1+ \tan \theta_1 \tan \theta_2 ) $$ $$ = \frac{p \cos \gamma}{a} \frac{ p \sin \gamma}{b} \left(1 -\frac{q^2}{p^2} \right) =\frac1{2ab} \sin(2\gamma) (p^2-q^2) $$ We already see the numerator $2 a b \cos \phi$. Now we will "foefel" ourselves to the solution by calculating $a^2-b^2$: $$ \begin{align} a^2-b^2&=p^2 \frac{\cos^2 \gamma}{\cos^2 \theta_1}-p^2 \frac{\sin^2 \gamma}{\cos^2 \theta_2} = p^2 \left[ \cos^2 \gamma\left(1+\tan^2(\theta_1)\right)- \sin^2 \gamma\left(1+\tan^2(\theta_2)\right) \right]\\ &=p^2 \left[ \cos^2 \gamma\left(1+\frac{q^2}{p^2} \tan^2 \gamma\right)- \sin^2 \gamma\left(1+\frac{q^2}{p^2} \cot^2 \gamma\right) \right]\\&=p^2 \cos^2 \gamma+q^2 \sin^2 \gamma-p^2 \sin^2 \gamma-q^2 \cos^2 \gamma=(p^2 -q^2) \cos (2\gamma) \end{align} $$ Putting everything together: $$ \cos \phi=\frac1{2ab} \sin(2\gamma) \frac{(a^2-b^2)}{\cos (2\gamma)}=\frac{a^2-b^2}{2ab} \tan(2\gamma) . $$ Rearranging gives your formula.