Find the condition that the second degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a rectangular hyperbola. Find the condition that the second degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a rectangular hyperbola.

I know that the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 \ne 0,h^2>ab,a+b=0$.But i dont know how to prove it.
I have read one proof on some website but it is proved by Eigen values and Eigen vectors.I could not understand that.
Is there any other method to prove this?Please help.
 A: The equations of the asymptotes, as they are perpendicular lines, can be written in general as:
$$
\alpha x+\beta y +\gamma=0
\quad\text{and}\quad
\beta x-\alpha y +\delta=0.
$$
The equation of the hyperbola can then be written as
$$
(\alpha x+\beta y +\gamma)(\beta x-\alpha y +\delta)=\varepsilon,
$$
where $\varepsilon\ne0$ if the hyperbola is non-degenerate.
This equation can be expanded, and comparing it with your
$ax^2+2hxy+by^2+2gx+2fy+c=0$ we find:
$$
a=-b=\alpha\beta,\quad 2h=\beta^2-\alpha^2,\quad
2g=\alpha\delta+\beta\gamma,\quad 2f=\beta\delta-\alpha\gamma,\quad
c=\gamma\delta-\varepsilon.
$$
The first equation is the same as your condition $a+b=0$, which then represents the perpendicularity of the asymptotes. From this it follows that $ab\le0$, and  condition $h^2>ab$ is then the same as stating that you cannot have $a=b=h=0$, for in that case the hyperbola would degenerate into a line.
The other condition for the hyperbola to be non-degenarate is $\varepsilon\ne0$, which can be written as $\gamma\delta-c\ne0$. But from the above relations we get
$$
\gamma=2{\beta g-\alpha f\over\alpha^2+\beta^2},\quad
\delta=2{\beta f+\alpha g\over\alpha^2+\beta^2},
$$
whence:
$$
\gamma\delta={4\alpha\beta(g^2-f^2)+4(\beta^2-\alpha^2)fg
\over(\alpha^2+\beta^2)^2}
={a(g^2-f^2)+2hfg
\over a^2+h^2}.
$$
Plugging the last result into $\gamma\delta-c\ne0$ we obtain
$$
a(g^2-f^2)+2hfg-(a^2+h^2)c\ne0,
$$
which is the same as your first condition, once you take into account that 
$a=-b$.
