Q:

Finding axis of ellipse described by $x=a\cos t+ h\sin t$,$ y=b\sin t + g\cos t$

Hi I am in need of help here for my project. Basically I have managed to obtain this form of equation. Example: $a=-181,h=33,b=185.9$ and $g=18.3$. When I plot it on a graphing program, it looks like a ellipse. Basically I need to find all the different parameters of the ellipse. Initially I thought that a general rotation would solve the problem.

Basically $x=A\cos t$ and $y=B\sin t$ would be a normal parametric equation of ellipse, and alpha is the angle I have rotated. Hence that would allow me to find the major and minor axis easily. However, as you can see solving the 2 equation actually yields 2 equations for $\alpha$ which are $\tan\alpha ={ -h\over b}$ and $\tan\alpha ={ g\over a}$.
Obviously this cannot be true. Does anyone know of what to do? Is there a more general equation of ellipse that I am unaware of? Please advise, thank you very much.

A:

One way to get an equation is to use  
$\begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix} a&h\\ g&b\end{pmatrix}\begin{pmatrix} \cos(t)\\ \sin(t)\end{pmatrix}$, which means that when the matrix is invertible
$\begin{pmatrix} \cos(t)\\ \sin(t)\end{pmatrix}=
{\begin{pmatrix} a&h\\ g&b\end{pmatrix}}^{-1}
\begin{pmatrix} x\\ y\end{pmatrix}$
then use $\cos^2(t)+\sin^2(t)=1$
That is $(bx-hy)^2+(-gx+ay)^2=(ab-gh)^2$ and in your example it is $(185.9x-33y)^2+(-18.3x-181y)^2=34251.8^2$
Now there are no linear terms, so you see that your ellipse is just a rotation of one centered at the origin. If you want you can get the semiaxes from this form by diagonalising the matrix
$\begin{pmatrix} b^2+g^2&-ag-bh\\ -ag-bh&a^2+h^2\end{pmatrix}$.

