<s>[INST] Equation of Conics The general equation of the conic section is : $ax^2+2hxy+by^2+2gx+2fy+c=0$,
where
$$\Delta=\begin{vmatrix}a&h&g\\h&b&f \\g&f &c\\
\end{vmatrix}$$
This equation can also be analysed to distinguish whether it is an equation of pair of straight lines, parabola, ellipse or hyperbola.

*

*If $\Delta=0$ and $h^2=ab$, it represents pair of straight lines

*If $\Delta \neq 0$ and $h^2=ab$, it represents a parabola

*If $\Delta \neq 0$ and $h^2<ab$, it represents an ellipse

*If $\Delta\neq 0$ and $h^2>ab$, it represents a hyperbola

So my question is
Can you represent a proof for why these are the relations for different shapes?
I would prefer a more analytic/geometric approach
 [/INST] Let's go the other way:
Any pair of lines has equation $k(a_1 x+b_1 y+c_1)(a_2 x+b_2 y+c_2)=0,$ which corresponds to the matrix being $\begin{pmatrix}a_1 \\b_1\\c_1\end{pmatrix}\begin{pmatrix}a_2&b_2&c_2\end{pmatrix},$ which is of rank $1$ and hence has zero determinant. Also all the 2 by 2 minors are zero in particular $h^2-ab=0.$
Now any parabola can be written as a multiple of  $\frac{(a_d x+b_d y+c_d)^2}{a_d^2+b_d^2}=(x-f_x)^2+(y-f_y)^2$ or $-k\frac{(b_d x-a_d y)^2+((-2b_d^2f_x)-2a_d^2f_x-2a_dc_d)x+((-2b_d^2f_y)-2a_d^2f_y-2b_dc_d)y+(b_d^2f_y^2+a_d^2f_y^2+b_d^2f_x^2+a_d^2f_x^2-c_d^2)}{a_d^2+b_d^2}=0.$ That is the first minor $\frac{k^2}{(a_d^2+b_d^2)^2}\begin{pmatrix}b_d^2&-a_db_d\\-a_db_d&a_d^2\end{pmatrix}$ vanishes or $h^2=ab.$
More generally any ellipse can be given as having equation proportional to $e^2\frac{(a_d x+b_d y+c_d)^2}{a_d^2+b_d^2}=(x-f_x)^2+(y-f_y)^2$ with the squared eccentricity $0<e^2<1$ and any hyperbola by the same equation with $e^2>1$. Making the first minor
$\frac{k^2}{(a_d^2+b_d^2)^2}\begin{pmatrix}a_d^2(e^2-1)-b_d^2&a_db_de^2\\a_db_de^2&b_d^2(e^2-1)-a_d^2\end{pmatrix},$ which has determinant proportional to $(e^2-1)(a_d^2+b_d^2)^2.$
</s>