Classification of conics based on general form

Question

I would like to ask questions as written. If we have an equation

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

how do we determine whether it is an ellipse, hyperbola, parabola, or none at all?

I have searched the internet and could not find any, unless they assume $B=0$ (including in MSE unless I missed some threads). Even worse, some say when $A=C$, it is immediately a circle without assuming $B=0$. Hence I did not trust the rest.

I know we also need to consider the sign of $B^2 - 4AC$.

Is there a complete "if ... then ...", case-by-case classification based on the above general form? A table, link, or source would be appreciated too. Thank you very much and stay safe.

EDIT: I have seen the Wikipedia page too. However, don't we still need to know whether it is degenerate or not before using the $B^2 - 4AC$ classification? This is part of my above question. By complete classification, I mean it also includes something like "If $A,B,C$ satisfies ... and ... or ..., then the curve is ...", where the hypothesis runs through all possibilities. Or, in other forms of classification, "It is ellipse if and only if $A,B,C$ satisfies ..." and so on for hyperbola and parabola.

Answer 1

To methodically solve this, you want to first eliminate the $D$ and $E$ terms; this can be done by replacing $x$ and $y$ with $x-a$ and $y-b$ for appropriately chosen $a, b$.

So we can now assume that our equation is of form $Ax^2+Bxy+Cy^2+F=0$. Factorise this as $$A\left(x+\frac{B}{2A}\right)^2+\left(C-\frac{B^2}{4A^2}\right)y^2+F=0.$$ Let $u = x+\frac{B}{2A}$, $v = y$, $G = C-\frac{B^2}{4A^2}$. Then we can write this as $$Au^2+Gv^2+F=0,$$ and the original function will have the same form as this conic, which is easy to check.

Answer 2

In $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ let $x=x'\cos{\theta}+y'\sin{\theta}, y=-x'\sin{\theta}+y'\cos{\theta}$

Expanding, the coefficient of $x'y'$ is $$-B\sin(\theta)^2+(2A-2C)\cos(\theta)\sin(\theta)+B\cos(\theta)^2=B\cos(2\theta)+(A-C)\sin(2\theta).$$

If $C\neq A$ are we can get this term to vanish by choosing $\tan2\theta=\frac{B}{C-A}.$ If $C=A$ we can choose $\theta$ such that $\cos2\theta=0.$

The point of this is that the discriminant is invariant and in each class of conic there's a rotated one with $B=0.$ And when $B=0,$ $B^2-4AC$ is determined by $A\cdot C.$

If $A=B=C=0;$ $Dx+Ey+F=0$ and we have a line, unless $D=E=0$ and we have the equation $F=0,$ which has no points unless $F=0$ and $0=0$ (which is satisfied for all points in the plane).

If the discriminant $B^2-4AC$ is zero and $B=0$ and either $C$ or $A$ is $\neq0,$ we have a parabola: $Ax^2+Dx+Ey+F=0$ or $Cy^2+Dx+Ey+F=0$ unless $D=E=0$ and we have $x^2=-1,0,1$ (no real points,a double line,two parallel lines).

If $B=0$ and the signs of $A, C$ are different, we have a hyperbola and can translate $Ax^2+Cy^2+Dx+Ey+F=0$ to $Ax^2+Cy^2+F'=0$ which can be brought into the form $x^2/a^2-y^2/b^2=1,$ or $x^2/a^2-y^2/b^2=0$ (a pair of lines).

If $B=0$ and the signs are the same, we can translate $Ax^2+Cy^2+Dx+Ey+F=0$ to $Ax^2+Cy^2+F'=0$ which can be brought into the form $x^2/a^2+y^2/b^2=1$ (and we have an ellipse (or circle if $a=b=r$)), or $x^2/a^2+y^2/b^2=0$ (which has one real point; the node of a pair of complex lines) or $x^2/a^2+y^2/b^2=-1$ (which has no real points).

We can rotate back to the original situation unless $B$ already was $0$ and have the same types.