$Ax^2+Bxy+Cy^2=1$ with $A=C=1$ and $B=2$ should be a parabola (because $B^2 = 4 AC$). Instead, it represents parallel lines. What went wrong?

Question

My calculus book says

The equation $Ax^2+Bxy+Cy^2=1$ produces a hyperbola if $B^2>4AC$ and an ellipse if $B^2<4AC$. A parabola has $B^2=4AC$.

When I set $B=2,A=C=1$, the equation becomes $x^2+2xy+y^2=1$, which is equivalent to $(x+y)^2=1$ or $y=-x\pm1$. This is not a parabola but two parallel lines. What went wrong?

Answer 1

You are right: a nondegenerate parabola doesn't admit an equation of the form $Ax^2+2Bxy+Cy^2=1$.

Indeed, if $B^2=AC\ne0$, we can multiply all coefficients by $A$, finding $$ A^2x^2+2ABxy+B^2y^2=B $$ that becomes $(Ax+By)^2=B$. If $B>0$ this factors as $$ (Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0 $$ which is a pair of parallel lines.

If $B<0$, the factorization is over the complex numbers and we get no real point.

If $B=0$, then either $A=0$ or $C=0$, so the equation becomes either $Cy^2=1$ or $Ax^2=1$, again a situation as before.

Only nondegenerate conics with a center admit an equation of the given form, which is obtained by translating the center of the conic in the origin.

Answer 2

And if you put $A=C=-1$ and $B=0$, what you get is the empty set, not an ellipse. The statement should read:

Let $A,B,C,D,E,F\in\mathbb R$. If the set$$\{(x,y)\in\mathbb{R}^2\,|\,Ax^2+Bxy+Cy^2+Dx+Ey+F=0\}$$is neither the empty set nor a degenerate conic, then it is a hyperbola if $B^2>4AC$, a parabola if $B^2=4AC$, and an ellipse if $B^2<4AC$.

Answer 3

$Ax^2+Bxy+Cx^2=1$ is symmetric under the $\pi$ rotation. But parabola has no centre of symmetry.

You need at least one linear term. Simplest example is $x^2-y=0. A=1,B=C=0.$