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

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?

• Parallel lines are a form of degenerate conic. Effectively, they represent the "parabola" you get by slicing a cone that has degenerated into a cylinder. – Blue Jan 19 '19 at 9:54

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.

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 $$B^2<4AC$$.

• Can you provide some $A,B,C$ for which $B^2=4AC$ and the set is not a degenerate conic? I am getting two parallel lines no matter what I try. – W. Zhu Jan 19 '19 at 10:17
• My answer was wrong and I've edited it. – José Carlos Santos Jan 19 '19 at 10:21

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

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