# Conic section classification

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.

• you only need to look at the coefficient matrix of quadratic terms, that $\left[\begin{array}{cc}1 & h\\ h &1\end{array}\right]$, when is it positive definite or have zero eigenvalues? – m-agag2016 Dec 12 '16 at 10:50
• I do not know what a positive definite matrix is, but I looked online right now and I found that, through Sylvester's criterion, I can say that a matrix is positive definite if all of the leading principal minors are positive. In this case I have the principal minor of order one positive $(1>0)$ while that of order 2 is $1-h^2$ which is strickly greater than zero only if $h<0$. Anyway, I cannot see how this helps :( – M.Giacchello Dec 12 '16 at 11:03
• For $h=1$, we have $(x+y)^2=5$. For $h=-1$, we have $(x-y)^2=3$. You may want to see here. – mathlove Dec 12 '16 at 11:04
• en.m.wikipedia.org/wiki/… – lab bhattacharjee Dec 12 '16 at 12:01

## 2 Answers

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!

A general conic in the affine plane has equation

$$ax^2 + by^2 +2hxy + 2gx+2fy+c = 0$$

but they're better though of as curves in the projective plane. In homogeneous coordinates:

$$ax^2 + by^2 +2hxy + 2gxz+2fyz+cz^2 = 0$$

The line at infinity has equation $z=0$, and the intersections with the line at infinity have

$$ax^2 + by^2 +2hxy = 0$$

Hyperbolae meet the line at infinity twice. Parabolae are tangent to the line at infinity. Ellipses miss the line at infinity. The number or roots of $ax^2 + by^2 +2hxy = 0$ classifies the conic.

Assuming that $a \neq 0$, $$\frac{x}{y} = \frac{-h\pm\sqrt{h^2-ab}}{a}$$ Assuming that $b \neq 0$, $$\frac{y}{x} = \frac{-h\pm\sqrt{h^2-ab}}{b}$$

We have a hyperbola is $h^2-ab > 0$, a parabola if $h^2-ab=0$ and an ellipse if $h^2-ab<0$.