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Suppose we consider the following hyperbola: $$x^2-2y^2+4xy-3x+5y+9=0$$

I wanted to find out something about this hyperbola in a question. That led me to think if I could try and find all the information I can.

We can find the center by solving the two equations:

$$\frac{\partial {\psi}}{\partial x} =0\,\tag1$$ $$\frac{\partial {\psi}}{\partial y} =0\,\tag2$$

where, $\psi\equiv x^2-2y^2+4xy-3x+5y+9$

This gives us the center as: $C\equiv\left(-\dfrac{4}{3},\dfrac{11}{12}\right)$

However when we want to find the axes of this hyperbola, the only method that I know of is to rotate the hyperbola, is to rotate the coordinate axes first by an angle $\theta$ such that, $$\tan2{\theta}=\frac{2h}{a-b}$$ for the general case of $$ax^2+by^2+2hxy+2gx+2fy+c=0$$ and then it would simply convert to the standard form of the hyperbola. Now we would be able to get back to our original coordinate system and get the equations of the transverse and conjugate axes. However this method, is clearly quite rigorous, especially for cases where some information is required for a general hyperbola.

So is there a more effective and less rigorous method? Maybe some playing around with the asymptotes? Maybe some calculus. I have had no progress so far.

Please note that even a method that's probably not short is welcome, provided it is something with a different flavor, and different approach.

I have searched StackExchange, but could only find an answer on the rotation of the axes and not what is asked here, therefore, I would not be linking it.

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  • $\begingroup$ The axes of a hyperbola (indeed any conic section) with your general equation has equation $$h^3y^2-abhy^2-bh^2xy+ah^2xy+ab^2xy-a^2bxy+2gh^2y-bfhy-afhy+b^2gy-abgy-h^3x^2+abhx^2-2fh^2x+bghx+aghx+abfx-a^2fx+g^2h-f^2h+bfg-afg=0.$$ In your example this factorises to: $\frac14(4y+8x-1)(12y-6x-13).$ $\endgroup$
    – Jan-Magnus Økland
    Commented Sep 13, 2020 at 13:28

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The axis directions are given by the eigenvectors of the quadratic form matrix – $x^2$ coefficient at top-left, $y^2$'s at bottom-right and the $xy$ coefficient is split between the two remaining entries: $$\begin{bmatrix}1&2\\2&-2\end{bmatrix}$$ In this case, the eigenvectors are $(-1,2)$ and $(2,1)$, so the axes radiate out in those directions from $C$.

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  • $\begingroup$ is the xy coefficient split equally between the two? $\endgroup$
    – Tesla's Coil
    Commented Sep 14, 2020 at 8:02
  • $\begingroup$ And is this valid for all hyperbolas? $\endgroup$
    – Tesla's Coil
    Commented Sep 14, 2020 at 8:03
  • $\begingroup$ And eigenvectors as I know them, can be obtained by solving (A-λI).X = O. Where, X is the eigen vector. Can we find eigenvectors in a better way, directly from the value of λ or otherwise? $\endgroup$
    – Tesla's Coil
    Commented Sep 14, 2020 at 9:08

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