I was given this equation $f(x,y) = x^2 + 4xy + y^2 - 2x + 2y + 1$
I'm trying to find the quadric surface when $f(x,y) = z$ and the quadric curve when $f(x,y) = 0$. I started with the quadric surface and set x to a constant and found that yz makes a parabola, set y to a constant and found that xz makes a parabola. But because there are mixed variables, I'm not sure what to make of it when I set z to a constant. I used Geogebra and Desmos to check my answer, I'm pretty sure the last part makes a hyperbola and the quadric curve is a hyperbola as well, but I don't know how to prove it.
$z = c : x^2 + 4xy + y^2 - 2x + 2y = c - 1$
What makes this a hyperbola? All the videos I found online doesn't go over examples with mixed variables, how I do a identify these when mixed variables are present?
EDIT: So I'm pretty sure the quadric surface is a hyperbolic paraboloid.
For any quadric, and this includes conics, you can rewrite the expression as a linear combination of squared linear terms and linear terms. This amounts to applying an affine transformation, which doesn’t change the nature of the quadric, to a coordinate system in which the equation has one of the standard forms that doesn’t include any cross-terms.
Taking first the level set $f(x,y)=0$, by repeatedly completing the square we can rewrite the equation as $$(x+2y-1)^2-3(y-1)^2+3=0.$$ This has the form $X^2-3Y^2+c=0$, which is a hyperbola.
We can analyze $f(x,y)-z=0$ in a similar fashion and can reuse most of the preceding work. We end up with $$(x+2y-1)^2-3(y-1)^2-(z-1) = 0.$$ This equation has the form $X^2-Y^2=Z$. If you don’t happen to recognize this as a hyperbolic paraboloid, you can analyze this simplified equation in the way that you describe in your question.
Effectively, what we’re doing here is computing the spectrum of the (homogenized) quadric, which uniquely determines its type.