In physic context one find the curve with parametrisation in t, $x=x_0\cos(t)$ and $y=y_0\cos(t+\varphi)$ with is an ellipse with equation
$$\left(\frac{x}{x_0\sin(\varphi)}\right)^2+\left(\frac{y}{y_0\sin(\varphi)}\right)^2-2\cos(\varphi)\left(\frac{x}{x_0\sin(\varphi)}\right)\left(\frac{y}{y_0\sin(\varphi)}\right)=1$$
I'd like to find a simple expression for the axes of that ellipse but I don't find: the associated quadratic form is
$$Q=\left(\begin{array}{cc} \left(\frac{1}{x_0\sin(\varphi)}\right)^2 & -\frac{\cos(\varphi)}{x_0y_0\sin^2(\varphi)} \\ -\frac{\cos(\varphi)}{x_0y_0\sin^2(\varphi)} & \left(\frac{1}{y_0\sin(\varphi)}\right)^2 \end{array}\right)$$
This quadratic form has the same eigen vectors as the following
$$Q'=\left(\begin{array}{cc} \frac{y_0}{x_0} & -\cos(\varphi) \\ -\cos(\varphi) & \frac{x_0}{y_0}\end{array}\right)$$
with caracteristic polynomial $\chi(\lambda)=\lambda^2-(\frac{x_0}{y_0}+\frac{y_0}{x_0})\lambda+\sin^2(\varphi)$ and so $\Delta=(\frac{x_0}{y_0}+\frac{y_0}{x_0})^2-4\sin^2(\varphi)$ wich is not sympathic.
Is there a simpler approach?
Let $$ Q=\begin{bmatrix}a & c\\ c & b\end{bmatrix} $$ be the quadratic form matrix of the ellipse $x^TQx=1.$ Let $\theta$ be a rotation which, when applied to the ellipse, will align the major and minor axes with the coordinate axes. Define the orthogonal matrix $$ R=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} =\cos\theta\begin{bmatrix} 1 & -\tan\theta\\ \tan\theta & 1\end{bmatrix}. $$ We determine $\tan\theta,$ which is minus the slope of one of the axes of the ellipse, by imposing the condition that $RQR^T$ be diagonal, that is, by setting the $(1,2)$ entry of $RQR^T$ to zero. This gives a quadratic equation for $\tan\theta$ in terms of $a,$ $b,$ and $c.$ It's slightly messy, but I don't think it can be made much simpler.
The result of the computation outlined above for your values of $a,$ $b,$ and $c$ is $$ \tan\theta=\frac{1}{2\cos\varphi} \left(\frac{x_0}{y_0} - \frac{y_0}{x_0} \pm \sqrt{\frac{x_0^2}{y_0^2} + \frac{y_0^2}{x_0^2} + 2 \cos(2\varphi)}\,\right). $$