Shape of level curves and Hessian determinant (Calculus 3) 
Consider the surface $$Q(x,y)=Ax^2+Bxy+Cy^2$$ and the Hessian determinant $4AC-B^2$.
How does the sign of the Hessian determinant determine the shape of $Q$'s level curves?
  When are the level curves ellipses?
  When are the level curves hyperbolas?

Progress so far:
Through some graphing, I'm pretty sure that ellipses occur when the Hessian determinant is $>0$ and hyperbolas occur when it is $<0$.
I can write this: $Q=A\left(\left(x+\frac{B}{2A}y\right)^2+\frac{4AC-B^2}{4A^2}y^2\right)$, which brings the determinant into the equation for $Q$, but I'm stuck here.
I suspect this has something to do with conic sections because it looks really similar. When you have a conic section in Cartesian form, the sections are ellipses when $B^2-4AC<0$ and hyperbolas when $B^2-4AC>0$: essentially the same as here. But I haven't found a connection to the level curves. Any help?
 A: The definition of the $k$ level curve is the set of points $(x, y)$ in the plane that satisfy $k = Q(x, y)$.
Introducing some new variables, we have
$$
\begin{align*}
Q(x, y) &= k\\
Ax^2 - Bxy + Cy^2 &= k\\
Ax^2 - Bxy + Cy^2 + Dx + Ey &= F, \tag{1}
\end{align*}
$$
where $D = E = 0$ and $F = k$.
Now we can use the fact you mentioned, namely that

When you have a conic section in Cartesian form, the sections are ellipses when $B^2 - 4AC < 0$ and hyperbolas when $B^2 - 4AC > 0$.

In fact, we do have a conic section in standard (Cartesian) form—it is the equation in $(1)$.
Therefore, we can say that the set of points $(x, y)$ in the plane that satisfy $(1)$ form an ellipse if $B^2 - 4AC < 0$ or a hyperbola if $B^2 - 4AC > 0$.
And, because the definition of the $k$ level curve is the set of points in the plane that satisfy $(1)$, the same is true of the level curve.
If the question is asking you to prove the fact you stated, then that's a different matter.
But as written, it looks like you have all the pieces, and just need to put them together.
