Plotting contours of the paraboloid $f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$

Question

I have the following function: $$f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$$ and I need to plot its contours.

My Attempt

I recognized this is a conic section given in its Cartesian form: $$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$ Since $B^2-4AC=16 - 4\times 4 = 0$ we know this is a parabola. Looking at the General Case of a parabola, I rewrite the contours in this way:

$$4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1=C \quad \quad \text{for } C\in\mathbb{R} $$ $$x_2^2+2x_2(1-2x_1) + 4x_1^2-4x_1+1 - C = 0$$ $$(x_2 + (1-2x_1))^2 - C = 0$$ This would mean that $$x_2 + 1 - 2x_1 = 0$$ is the directrix of the parabola, i.e. $x_2 = 2x_1 - 1$. Here's a plot of that:

However, I have no idea how to proceed from here. I am not even sure that what I've done makes sense, to be honest.

Answer 1

$(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=\sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-\sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.

The plot must consist mostly in two parallel lines.

Answer 2

Besides the discriminant, you also have to examine the full $3\times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.