# 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.

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 $a=-b$.