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?
On
I give here another parametrization, a little harder and less practical but somewhat enlightening for the beginners. The focus is $F=\left(\dfrac{-7}{40},\dfrac{11}{40}\right)$ and the directrice is $4x+12y-1=0$ then if $(x,y)$ is any point of the parabola then its distance to the focus and to the directrice are equal by definition of parabola. This gives the equation $$(40x+7)^2+(40y-11)^2=10(4x+12y-1)^2$$ Because $10=3^2+1$ we have by the known parametrization of solutions of the equation $x^2+y^2=z^2+w^2$ $$\begin{cases}40x+7=rX+sY\\40y-11=rY-sX\\12x+36y-3=rX-sY\\4x+12y-1=rY+sX\end{cases}$$ from which $$\begin{cases}52x+36y+4=2rX\\4x+52y-12=2rY\\28x-36y+10=2sY\\4x-28y+10=2sX\end{cases}$$ so $$\frac{13x+9y+1}{x+13y-3}=\frac{2x-14y+5}{14x-18y+5}=t$$ Solving this system we get for $t$ real distinct of $1$ $$x=\frac{-11t^2+86t-59}{200(t-1)^2}$$
$$y=\frac{47t^2-62t+63}{200(t-1)^2}$$ This parametrization is equivalent to the simpler one suggested by the O.P.
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 $$ \begin{cases} x = x(t) = \frac{1}{8}\left(2t + t^2 - 2 \right), \\ y = y(t) = \frac{1}{8}\left(-2t + 3t^2 + 2 \right), \\ t \in \mathbb{R}. \end{cases} $$