How to really prove a curve with b^2-4ac>0 and a positive discriminant for all y in R is a hyperbola?

Question

I'm doing Exercise 1.1.18.(2) in Algebraic Geometry: A Problem Solving Approach (Garrity, et al.).

For a polynomial $P(x,y) = ax^2 + bxy + cy^2 + dx + ey + h$, suppose $b^2-4ac > 0$. If {$y | \Delta_x(y) \ge 0$} $= \mathbb{R}$, show that $V(P)$ is a hyperbola opening left and right.

I know that $\Delta_x(y) = (b^2-4ac)y^2 + (2bd-4ae)y + (d^2-4ah)$, but we haven't yet defined hyperbolas so I'm not sure what to do here exactly. With regards to hyperbolas, the text has stated that conics that have the symmetry and boundedness properties of an example curve $C =$ {$(x,y) \in \mathbb{R}^2 : x^2 - y^2 - 4 = 0$} (symmetric about both the $x$- and $y$-axes and bounded in the positive and bounded in one dimension so as to consist of two connected components) and look like $C$ are called hyperbolas. This isn't a clear definition, and I'm not sure if this is an exhaustive list of these types of conics, but here is what I did:

If {$y\Delta_x(y)\ge 0$} $= \mathbb{R}$, then $V(P)$ is unbounded in the positive and negative $y$-directions. So $V(P)$ is not an ellipse or a parabola, therefore it must be a hyperbola.

This feels overly simplistic. What am I missing?

Answer 1

The general hyperbola is a rotated and translated version of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or its conjugate $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$. This equation can be written $(\frac{x}{a}-\frac{y}{b})(\frac{x}{a}+\frac{y}{b})=1.$ Remembering that multiplying the equation by a non-zero constant doesn't change the hyperbola, a general equation (the image rotated and translated multiplied by a constant) will be of the form $\text{asympotote}_1\cdot \text{asympotote}_2=\text{constant}.$ Let's find out when the general equation is of this form.

Solving the system

eq:a*x^2 + b*x*y + c*y^2 + d*x + e*y + h;
solve([diff(eq,x),diff(eq,y)],[x,y]);

in maxima CAS, the center, provided $b^2-4ac\neq 0$ (i.e. not parabola or degenerate), is

$(x,y)=(\frac{2cd-be}{b^2-4ac},-\frac{bd-2ae}{b^2-4ac})$

Translating the equation moving the center to the origin

ratsimp(a*(x+(2*c*d-b*e)/(b^2-4*a*c))^2 + b*(x+(2*c*d-b*e)/(b^2-4*a*c))*(y-(b*d-2*a*e)/(b^2-4*a*c)) + c*(y-(b*d-2*a*e)/(b^2-4*a*c))^2 + d*(x+(2*c*d-b*e)/(b^2-4*a*c)) + e*(y-(b*d-2*a*e)/(b^2-4*a*c)) + h);

we see that, under the assumption $b^2-4ac\neq 0$, the original equation can be rewritten

$$a(x-\frac{2cd-be}{b^2-4ac})^2 + b(x-\frac{2cd-be}{b^2-4ac})(y+\frac{bd-2ae}{b^2-4ac}) + c(y+\frac{bd-2ae}{b^2-4ac})^2 + h+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0$$ or $$a(x-\frac{2cd-be}{b^2-4ac})^2 + b(x-\frac{2cd-be}{b^2-4ac})(y+\frac{bd-2ae}{b^2-4ac}) + c(y+\frac{bd-2ae}{b^2-4ac})^2+\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&h\end{vmatrix}}{\begin{vmatrix}a&b/2\\b/2&c\end{vmatrix}}=0$$

Disregarding this last term $\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&h\end{vmatrix}}{\begin{vmatrix}a&b/2\\b/2&c\end{vmatrix}}$ it represents two lines (or more degenerate). These are the asymptotes and the conic is a hyperbola when representing a real line pair, and again, the discriminant $D=b^2-4ac > 0$ is the criterion for that, since the actual lines are given by $a((x-\frac{2cd-be}{b^2-4ac})+(y+\frac{bd-2ae}{b^2-4ac})(b+\sqrt{D})/2a)((x-\frac{2cd-be}{b^2-4ac})+(y+\frac{bd-2ae}{b^2-4ac})(b-\sqrt{D})/2a)=0.$

In conclusion the equation represents a hyperbola $$a((x-\frac{2cd-be}{b^2-4ac})+(y+\frac{bd-2ae}{b^2-4ac})(b+\sqrt{D})/2a)((x-\frac{2cd-be}{b^2-4ac})+(y+\frac{bd-2ae}{b^2-4ac})(b-\sqrt{D})/2a)\\+\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&h\end{vmatrix}}{\begin{vmatrix}a&b/2\\b/2&c\end{vmatrix}}=0$$ when $b^2-4ac > 0$ and $\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&h\end{vmatrix}\neq 0.$