# Finding the conjugate hyperbola

Assume we are given a general proper hyperbola $a_{11}x^2 + 2 a_{12} x y + a_{22} y^2 + 2 a_{13}x +2 a_{23}y + a_{33} = 0$ with $D =\det \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix} < 0$ and $A = \det \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \neq 0$.

Is there an easy way to find the implicit description of the conjugate hyperbola?

I could translate and rotate the hyperbola and obtain a canonical representation of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$, find the conjugate of the canonical one and the transform back. However, I'm looking for a more straightforward way to achieve this.

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I think you're missing the linear terms in the equation for the general proper hyperbola. –  Andrew Aug 14 '12 at 13:48
Naturally I meant the conjugate... Sorry for the mistake –  Dror Aug 14 '12 at 14:56

Using the right translation and rotation one can obtain for a general hyperbola, as given in the OP, a canonical congruent hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$. The canonical conjugate is then given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \mp 1$, and can be transformed back (using inverse transformations) to obtain a general form of the conjugate hyperbola of the one we started with. Doing some algebra its the implicit representation of the conjugate hyperbola is given by $a_{11}x^2 +2a_{12}xy+a_{22}y^2 +2a_{13}x+2a_{23}y+a_{33} −2\frac{A}{D}$ with $D,A$ are the determinants given in the OP.