How to get a implicit equation of this ellipse If I give you the curve $\gamma (t) = ( \cos (t), \cos(t+a))$, how can I obtain an implicit equation? or which change of basis can I do to get the canonical implicit form?
 A: Basically the same method as described in this answer: Expand $y=\cos(t+\alpha) = \cos t\cos\alpha-\sin t\sin\alpha$ and solve for $\sin t$ (disregard possible zero denominators for now). Substitute $x=\cos t$ into the result, then substitute both into $\cos^2t+\sin^2t=1$, rearrange and simplify.
A: This is related to MSE question 2881242. 
Define variables $ u := \cos(t), \, v := \cos(t+a), \, w := \cos(a). \,$ The implicit equation is
$\, u^2 + v^2 + w^2 - 2 u v w = 1. $
A: The canonical form is $A x^2 + B y^2 +2 C x y = 1$
Substitute $x=\cos(t)$ and $y = \cos(a)\cos(t)-\sin(a)\sin(t)$ into the above and try to eliminate all coefficients of $\sin(t)$ and $\cos(t)$. 
It helps to use the substitutions $\cos^2(t) = \frac{\cos(2 t)+1}{2}$, $\sin^2(t) = \frac{1-\cos(2 t)}{2}$ and $\sin(t)\cos(t) = \frac{\sin(2 t)}{2}$
You will get in the end the following
$$  \begin{multline} \left( B \cos^2(a) + C \cos(a) + \frac{A-B}{2} \right) \cos(2 t) - \Bigl( B \sin(a)\cos(a)+C \sin(a) \Bigr) \sin(2 t)+ \\  + \left( C \cos(a) + \frac{A+B}{2} \right) = 1 \end{multline} $$
Now we eliminate all coefficients of $\sin(2 t)$ and $\cos(2t)$ to convert the above into a system of equations
$$\begin{aligned}
  B \cos^2(a) + C \cos(a) + \frac{A-B}{2} & = 0 \\
  B \sin(a)\cos(a)+C \sin(a) & = 0 \\
  C \cos(a) + \frac{A+B}{2} & = 1 
\end{aligned} $$
Solve the above for $A$, $B$ and $C$.
