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This question already has an answer here:

If I have the coefficients of the following equation:

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?

i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.

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marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander Henderson Aug 6 '18 at 1:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward. $\endgroup$ – amd Jul 24 '18 at 5:59
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First find the centre of the conic. This is the point $(u,v)$ such that the equation of the conic can be rewritten as $$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$ For the conic to be a hyperbola, the quadratic part has to factor into distinct linear factors over $\Bbb R$: $$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$ Then the asymptotes are $$R_i (X-u)+S_i(Y-v)=0$$ ($i=1$, $2$).

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  • $\begingroup$ Looks good. I just need to figure out the degenerate cases and how to handle them... $\endgroup$ – Justin Jul 24 '18 at 18:52
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The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as: $$ (ax+by+c)(dx+ey+f)=g. $$ You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $A\dots F$ to obtain $a\dots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $A\ne0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

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In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2\times2$ symmetric.

It can be centered with

$$(x+A^{-1}b)^TA(x+A^{-1}b)-b^TA^{-T}A^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TP\Lambda P^{-1}y+c'=z^T\Lambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$\lambda_uu^2-\lambda_vv^2=(\sqrt{\lambda_u}u+\sqrt{\lambda_v}v)(\sqrt{\lambda_u}u-\sqrt{\lambda_v}v)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^{-1}z-A^{-1}b.$$

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The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = \begin{bmatrix}A&\frac B2&\frac D2\\\frac B2&C&\frac E2\\\frac D2&\frac E2&F\end{bmatrix}$$ and the line at infinity is $\mathscr l = [0,0,1]^T$. Let $\mathscr l_\times$ be the “cross product matrix” of $\mathscr l$. We can compute the hyperbola-line intersection by finding a value of $\alpha$ that makes $$\mathscr l_\times^TQ\mathscr l_\times+\alpha\mathscr l_\times = \begin{bmatrix}C&-\alpha-\frac B2&0\\\alpha-\frac B2&A&0\\0&0&0\end{bmatrix}$$ a rank-one matrix. This occurs when the principal $2\times2$ minor vanishes, which leads to a quadratic equation in $\alpha$ with solutions $\pm\frac12\sqrt{B^2-4AC}$. Taking the positive root gives the matrix $$\begin{bmatrix}C&-\frac12\left(B+\sqrt{B^2-4AC}\right)&0 \\ -\frac12\left(B^2-\sqrt{B^2-4AC}\right)&A&0\\0&0&0\end{bmatrix}.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $C\ne0$, then the two points are $\mathbf p_1 = \left[C,-\frac12\left(B+\sqrt{B^2-4AC}\right),0\right]^T$ and $\mathbf p_2 = \left[C,-\frac12\left(B-\sqrt{B^2-4AC}\right),0\right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $\mathscr m_i = Q\mathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]\mathscr m_i=0$, or $[x,y,1]Q\mathbf p_i=0$.

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