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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.

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  • $\begingroup$ Hint: the first two conditions tell you that the equation represents a nondegenerate hyperbola. The last is related to the desired relation between its asymptotes. $\endgroup$ – amd May 29 '18 at 2:50
  • $\begingroup$ See en.m.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee May 29 '18 at 13:25
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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 $a=-b$.

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