Parametric Coordinates of an Ellipse through skewed Cartesian coordinates

Question

I wanted to find parametric coordinates of an ellipse purely in terms of trigonometric functions. For this I began with finding counterparts of $\sin$ and $\cos$ functions in skewed Cartesian coordinates and termed them as g(x) and f(x), respectively. Since $\cos(x),\sin(x)$ are the parametric coordinates of circle, I expected an ellipse upon plugging in $(f(x),g(x))$. With the help of Sine rule I established (here $\phi$ is the acute angle Y-axis makes with X-axis) $$f(x)=\frac{\sin(\phi-\theta)}{\sin(\phi)} $$ $$ g(x)=\frac{\sin(\theta)}{\sin(\phi)} $$ I did receive an ellipse as output, but with the major axis aligned to the line $y=x$ or $y=-x$ for all values of $\phi$, in contrast I expected it align with the line $y=\tan(\phi)$.I would like to know why this happened and what manipulation would correct it.

Answer 1

I don't know what kind of reasoning you made to write those equations, but you might have been misled by the circumstance that the parameter $\theta$ in the parametric equations of an ellipse is not the same as the polar angle. Better then to use letter $t$ to represent that parameter.

In general, any parametric equation of the form $$\vec p(t)=\vec\alpha\cos t+\vec\beta\sin t,$$ where $\vec\alpha$ and $\vec\beta$ are two constant vectors, represents an ellipse centred at the origin, having $\vec\alpha$ and $\vec\beta$ as conjugate semi-diameters.

Your equations are of that kind, with $$ \alpha=(1, 0),\quad \beta=(-\cot\phi, \csc\phi) $$ and that explains why they lead to an ellipse. On the other hand, that ellipse passes through $(\pm1,0)$ and $(0,\pm1)$ and that means, by symmetry, that the axes of the ellipse are the lines bisecting the quadrants.

This can also be found from the values of $t$ leading to the vertices of the ellipse, which can be computed from:

$$ \tan 2t={2\vec\alpha\cdot\vec\beta \over|\vec\alpha|^2-|\vec\beta|^2}=\tan\phi, \quad\text{that is:}\quad t={\phi\over2},\quad t={\phi+\pi\over2}. $$ Plugging this into your equations one gets in fact $y(t)=\pm x(t)$.