EXERCISE: Let $h\in\mathbb{R}$ and consider the conic section $C:x^2+2hxy+y^2-h-4$. Are there any values for which $C$ is a parabola? What kind of conic section is $C$ for $h=1$? And for $h=-1$?
SOLUTION: I investigate the determinants of the matrix of the quadratic equation $B$ and that of the matrix of the quadratic form $A$: \begin{equation} det(B)=\begin{vmatrix} 1 & h & 0 \\ h & 1& 0\\0&0&-h-4 \end{vmatrix} = (-h-4)\begin{vmatrix}1&h\\h&1\end{vmatrix}=(-h-4)(1-h^2)=h^3+4h^2-h-4 \end{equation} \begin{equation} det(A)=\begin{vmatrix}1&h\\h&1\end{vmatrix}=1-h^2 \end{equation} In order for $C$ to be a parabola, $det(B)$ needs to be non-zero, hence $h\neq\pm1\wedge h\neq-4$ (as $h^3+4h^2-h-4=(h+1)(h-1)(h+4)$), and $det(A)$ needs to be zero. Since the second condition implies $h=\pm1$, $C$ can never be a parabola.
Now, if $h=\pm1$, $B$ is singular, so that $C$ is a degenerate conic. In order to claffify what type of degenerate conic I have to inspect the determinant of $A$ but, by substituting $h=\pm1$ I get the zero. I have to conclude that for such values $C$ is not a conic or what? Have I made some mistakes?
Thank you all for you help.
I found that both $det(B)=det(A)=0$ hence, as suggested by @mathlove, I have to inspect the determinant of the matrix $C$, which is: \begin{equation}det(C)=\begin{vmatrix}a_{11}&a_{13}\\a_{13}&a_{33}\end{vmatrix}+\begin{vmatrix}a_{22}&a_{23}\\a_{23}&a_{33}\end{vmatrix}\end{equation} So, for $h=1$ I get \begin{equation}det(C)=\begin{vmatrix}1&0\\0&-5\end{vmatrix}+\begin{vmatrix}1&0\\0&-5\end{vmatrix}=-10<0\end{equation} and thus two real-parallel lines.
For $h=-1$ \begin{equation}det(C)=\begin{vmatrix}1&0\\0&-3\end{vmatrix}+\begin{vmatrix}1&0\\0&-3\end{vmatrix}=-6<0\end{equation} that is, again, a couple of parallel lines.
NOTE: I answered my own question just to close the topic. Thank you very much to everyone and especially to mathlove!