When my professor was going through coninc sections of the real plane he didn't really mention why such a curve has no real points iff the determinant of the matrix associated to the conic, A, without the first row and column is positive (this I get it, of course, it has to be an ellipse) and a(1,1) • det A > 0.
The second part is the one that messes me up: I've been trying to understand why it works this way for the last three days, unsuccessfully. Could someone show me a proof? My books don't even talk about it and I haven't found much on the web so far.
Thanks in advance!
Here is something slightly different from your criterion.
Using homogeneous coordinates $(x,y,t)$:
$$X^TAX=\begin{pmatrix}t&y&x \end{pmatrix}\begin{pmatrix}f& e&d\\e&c&b\\d&b&a \end{pmatrix}\begin{pmatrix}t\\y\\x \end{pmatrix}\tag{1}$$
(think to $t=1$ for "ordinary points")
Saying that this conic has no real points is the same as saying it remains always positive or always negative. See appendix for that.
Let us assume, up to a multiplication by $-1$ of matrix $A$ that we are in the first case ($\forall X, \ X^TAX > 0$). It means that $A$ is a positive definite matrix.
A classical necessary and sufficient condition for a matrix to be positive definite is Sylvester's criterion: either that:
$$f>0 \ \ \ \& \ \ \ \begin{vmatrix}f& e\\e&c \end{vmatrix}>0 \ \ \ \& \ \ \ \begin{vmatrix}f& e&d\\e&c&b\\d&b&a \end{vmatrix}>0$$
or / and (going from bottom right to upper left):
$$a>0 \ \ \ \& \ \ \ \begin{vmatrix}c& b\\b&a \end{vmatrix}>0 \ \ \ \& \ \ \ \begin{vmatrix}f& e&d\\e&c&b\\d&b&a \end{vmatrix}>0$$
You see that we are not far from your criterion...
Appendix: By contradiction, assume that for certain $X_1, X_2 \in \mathbb{R^3}$, we have simultaneously
$$X_1^TAX_1<0 \ \ \text{and} \ \ X_2^TAX_2>0$$
then, setting $X=\lambda X_1+X_2$, expanding
$$X^TAX=(\lambda X_1+X_2)^TA(\lambda X_1+X_2)$$
$$=(X_1^TAX_1)\lambda^2 + 2(X_1^TAX_2)\lambda + X_2^TAX_2$$
The discriminant of this quadratic in $\lambda$
$$\Delta=4((X_1^TAX_2)^2-(X_1^TAX_1)(X_2^TAX_2))$$
being $>0$, there exists a real root $\lambda$ giving a solution $X \in \mathbb{R^3}$ to $X^TAX=0$.