Cfrgtkky

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
$tan 2gamma = frac{2ab cos phi}{a^2 - b^2}$.

Can somebody prove this please?

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:
$$
begin{pmatrix}
x\
y
end{pmatrix}
=
begin{pmatrix}
cosgamma & -singamma\
singamma & cosgamma
end{pmatrix}
begin{pmatrix}
p cos u\
q sin u
end{pmatrix}
=
begin{pmatrix}
p cos ucosgamma-q sin gamma sin(u)\
p cos usingamma+q cos gamma sin(u)
end{pmatrix}
$$
$$
=
begin{pmatrix}
p cos gamma left(cos u-frac{q sin gamma}{p cos gamma} sin u right)\
p sin gamma left(cos u+frac{q cos gamma}{p sin gamma} sin u right)
end{pmatrix}
$$

Now say
$$
tantheta_1=frac{q}{p} tan gamma,
$$
$$
tantheta_2=-frac{q}{p} cot gamma.
$$
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:
$$
cosphi=cos(theta_2- theta_1)=cos theta_2 cos theta_1+sin theta_2 sin theta_1=cos theta_2 cos theta_1 (1+ tan theta_1 tan theta_2 )
$$
$$
= frac{p cos gamma}{a} frac{ p sin gamma}{b} left(1 -frac{q^2}{p^2} right) =frac1{2ab} sin(2gamma) (p^2-q^2)
$$
We already see the numerator $2 a b cos phi$. Now we will "foefel" ourselves to the solution by calculating $a^2-b^2$:
$$
begin{align}
a^2-b^2&=p^2 frac{cos^2 gamma}{cos^2 theta_1}-p^2 frac{sin^2 gamma}{cos^2 theta_2} = p^2 left[ cos^2 gammaleft(1+tan^2(theta_1)right)- sin^2 gammaleft(1+tan^2(theta_2)right) right]\
&=p^2 left[ cos^2 gammaleft(1+frac{q^2}{p^2} tan^2 gammaright)- sin^2 gammaleft(1+frac{q^2}{p^2} cot^2 gammaright) right]\&=p^2 cos^2 gamma+q^2 sin^2 gamma-p^2 sin^2 gamma-q^2 cos^2 gamma=(p^2 -q^2) cos (2gamma)
end{align}
$$
Putting everything together:
$$
cos phi=frac1{2ab} sin(2gamma) frac{(a^2-b^2)}{cos (2gamma)}=frac{a^2-b^2}{2ab} tan(2gamma) .
$$
Rearranging gives your formula.