Parametric equation for an ellipse from these parametric equations I have an ellipse defined as the following parametric equations:
$$x(t)=a\cos{(t+\phi_1)}$$
$$y(t)=b\cos{(t+\phi_2)}$$
$$0 \le t < 2\pi$$
I would like to know the length of the semi-major and semi-minor axes, and rotation angle, of this ellipse (in terms of $a$, $b$, $\phi_1$, and $\phi_2$). I have tried to transform these equations into the standard parametric form of an ellipse as given in this answer, but I have failed.
If we define $ϕ3=ϕ2+π/2$ then we have $x(t)=a\cos t\cosϕ_1−a\sin t\sinϕ_1$
and $y(t)=b\sin t\cosϕ_3+b\cos t\sinϕ_3$. Comparing coefficients with the linked answer, we then have $a\cosϕ_1=R_x\cosθ$, $a\sinϕ_1=R_y\sinθ$, $b\sinϕ_3=R_x\sinθ$, and $b\cosϕ_3=R_y\cosθ$.
This is where I'm stuck though, because I keep going in circles hereon out. I know that it should be solvable because I have four equations for three unknowns ($θ$, $R_x$, and $R_y$).
 A: The solution seems messy. One approach to finding the major and minor axes is to find the critical points of the squared-distance-to-origin function
$$
f(t):=[a\cos t]^2 + [b\cos(t+\beta)]^2 ;\tag1
$$
to simplify a bit I've reparametrized so that $\phi_1=0$ and I've written $\beta$ in place of $\phi_2-\phi_1$. The critical points $t^*$ and $t^{**}$ between $0$ and $\pi$ where $f'(t)=0$ will be $\frac\pi2$ apart and will correspond to the semi-major and semi-minor axes. Their squared lengths will be obtained by plugging $t^*$ and $t^{**}$ back into (1). The rotation angle will be the inclination of the line segment from $(0,0)$ to $(x(t^*), y(t^*))$ [or $(x(t^{**}), y(t^{**}))$, take your pick].
To find the critical points, differentiate (1) and simplify:
$$
\begin{align}
f'(t)&=-2a^2\cos t\sin t - 2b^2 \cos(t+\beta)\sin(t+\beta)\\
&=-\left(a^2\sin(2t) + b^2\sin(2t+2\beta)\right)\\
&=-\left(\sin(2t)[a^2+b^2\cos(2\beta)]+\cos(2t)[b^2\sin(2\beta)]\right)
\end{align}
$$
so the critical points satisfy the equation
$$\tan(2t)=-\frac{b^2\sin(2\beta)}{a^2+b^2\cos(2\beta)}.
$$
