# 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.

• 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.
Looking first at the conjugate hyperbolas in canonical position ${x^2\over a^2}-{y^2\over b^2}=\pm 1$, it’s evident from inspection that the determinants of the corresponding coefficient matrices have determinants with the same magnitude but opposite signs. Rotation and translation doesn’t change these determinants, so this property holds for any conjugate pair. Now, the hyperbolas of the one-parameter family $a_{11}x^2 + 2 a_{12} x y + a_{22} y^2 + 2 a_{13}x +2 a_{23}y + k = 0$ share a common pair of asymptotes, so finding the conjugate of the original hyperbola becomes a matter of finding the value of $k$ that produces a determinant of $-A$. If you work though the algebra, you’ll find that $k=a_{33}-2\frac AD$, in agreement with Dror’s answer.
If $k=k_0=\frac AD -a_{33}$, then the hyperbola will be degenerate and consist of the asymptotes. So, if you happen to know the asymptotes and can compute $k_0$ (say, by multiplying their equations together and adjusting coefficient to match), then the constant term of the conjugate hyperbola will be $2k_0-a_{33}$.