$Ax^2+Bxy+Cy^2=1$ 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 $B^2&gt;4AC$ and
  an ellipse if $B^2&lt;4AC$. 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?
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 $B&gt;0$ this factors as
$$
(Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0
$$
which is a pair of parallel lines.

If $B&lt;0$, 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.
$Ax^2+Bxy+Cx^2=1$ is symmetric under the $\pi$ rotation.  But parabola has no centre of symmetry.

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