The question is how to use matrices to determine the type of curve and its characteristics. The curve is
5x^2-4xy+2y^2-24 = 0
Any help is appreciated
Apply a coordinate rotation to the (x, y) coordinate system:
$\displaystyle \left ( \begin{matrix} x \\ y \end{matrix} \right ) = \left ( \begin{matrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{matrix} \right ) \left ( \begin{matrix} x' \\ y' \end{matrix} \right )$
such that when the rotation is applied the coefficient of x'y' is zero. Then you can apply the discriminant test to the equation in x' and y' to determine the kind of curve.
For example, in this case I get the coefficient of the x'y' term to be
$\displaystyle -8~cos^2(\theta) + 6~sin(\theta)~cos(\theta) + 4$
The solution here is nasty and I could only do it numerically. I got $\displaystyle \theta \approx 0.46365~rad$. The equation becomes
$\displaystyle 6x'^2 + y'^2 - 24 = 0$
which is clearly an ellipse.
-Dan