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?

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 $C$.
