<s>[INST] Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$ 
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$

My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
 & \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
 & \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -2y=0\\
 & \iff (3x-y+1)^2-2(x+y)=0.\end{split}
So this look like some crazy parabola, right? But I don't know how to derive parametric equation of this?
 [/INST] One possible parametrization can be obtained by setting $u = 3x-y+1$ and $v = 2(x + y)$, so that
$$
\begin{cases}
u = 3x - y + 1 \\
v = 2(x + y)
\end{cases} \Longleftrightarrow
\begin{cases}
x = \frac{1}{8} \left(2u + v - 2 \right)\\
y = \frac{1}{4} \left(-2u + 3v + 2 \right)
\end{cases}
$$
and $9x^2 +y^2-6xy+4x-4y+1=0 \Leftrightarrow u^2 = v$. New parabola can be trivially parametrized as $(u, v) = (t, t^2)$ for $t \in \mathbb{R}$, which leads to
$$
\begin{cases}
x = x(t) = \frac{1}{8}\left(2t + t^2 - 2 \right), \\
y = y(t) = \frac{1}{8}\left(-2t + 3t^2 + 2 \right), \\
t \in \mathbb{R}.
\end{cases}
$$
</s>