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My frustration started after hours of searching failed to turn up a formula for the vertex of a parabola in the general form

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

As is already well known, the discriminant $\Delta=B^2-4AC$ can be used to diagnose the type of conic one has if it it given in the general form. Other useful formulae I found in the midst of searching include the formula for the slope of the principal axis:

$$\tan\;\theta=\frac{B}{A-C+(\mathrm{sgn}\;B)\sqrt{B^2+(A-C)^2}}$$

the eccentricity (intended only for elliptic or hyperbolic cases)

$$\varepsilon=\left(\sqrt{\frac12-\frac{(\mathrm{sgn}\;\Delta)(A+C)}{2\sqrt{B^2+(A-C)^2}}}\right)^{-1}$$

and the coordinates of the center for a central conic (I won't list it here so you can figure it out or find it yourself ;) ), but there were no formulae given for finding the coordinates of the vertex of a parabola.

Of course, for the parabola case, I could rotate axes, find the vertex through completing the square, and then rotate back, but I was hoping somebody already went through the trouble of deriving a formula so I don't have to reinvent the wheel.

On to my query: is there a comprehensive compilation (book, article) somewhere of formulae related to dealing with conic sections? The implicit Cartesian form is what I'm currently dealing with, but lists relating to other forms (polar, parametric) are welcome as well.

J. M. ain't a mathematician
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    My suggestion is the book "Analytic geometry: a vector approach", by Charles Wexler. There is a very good chapter on conic sections, proving each passage. It's very clear. – Ronaldo Sep 14 '10 at 23:32
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    I remember there used to be a book by S.L.Loney which was filled with cartesian geometry results about conic sections. Don't recollect the name of the book though. – Aryabhata Sep 15 '10 at 00:18
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    I don't know what the principal axis is but if you know its slope you should be able to find the vertex by finding the point whose tangent line has the appropriate slope. – Qiaochu Yuan Sep 15 '10 at 03:58
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    Qiaochu: it's the line of symmetry for a conic (one of the lines of symmetry for a central conic). From $\tan;\theta$ you can compute the sine and cosine, and thus rotate the conic to a form where the more usual formulae apply (algebraically, it removes the cross term $xy$). Done by hand, I can easily find the vertex of a parabola, but this is meant for computer implementation. – J. M. ain't a mathematician Sep 15 '10 at 04:15
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    The thing with the parabolic case... as I said, I'd prefer something that doesn't involve rotate, complete the square, undo rotation on vertex thus found since this is intended for a dumb computer (without geometric insight) to do. – J. M. ain't a mathematician Sep 15 '10 at 04:19
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    @J. M.: then all you have to do is find the point whose tangent line has slope -1/tan theta. Right? – Qiaochu Yuan Sep 15 '10 at 04:20
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    @Qiaochu: Ah, that's where I got stuck when I tried to derive it myself. From taking partial derivatives of the standard form I wound up with $\frac{Bx+2Cy+E}{2Ax+By+D}=-\mathrm{cot};\theta$, solving that as a simultaneous equation with the original quadratic resulted in something nightmare-ish. I can't help but feel there's a more elegant way of going about it... – J. M. ain't a mathematician Sep 15 '10 at 04:41
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    How about reducing this quadratic form to canonical form (by means of non-singular linear transform), where finding of any special point is trivial, then transforming the found point back to original domain via inverse transform? Lagrange's reduction is straightforward and simple. – mbaitoff Jan 15 '11 at 19:54
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    Unless I or my CAS messed up my computation, the vertex of the parabola is at $$\frac1{(16A^4B+8A^2B^3+B^5)D-2(16A^5+8A^3B^2+AB^4)E}\cdot\\begin{pmatrix}16A^4DE-4A^3BD^2+(8A^3B+AB^3)E^2-(16A^4B+8A^2B^3+B^5)F\2A^2B^2E^2-2AB^3DE-4(2A^4+A^2B^2)D^2+2(16A^5+8A^3B^2+AB^4)F\end{pmatrix}$$ although there might be an easier way to write this. – MvG Aug 03 '13 at 01:25
  • @MvG, if memory serves, what I obtained the last time I looked into this was just as horrendous; thanks, nevertheless. – J. M. ain't a mathematician May 02 '15 at 03:17
  • I have added an answer to another question; this gives the foci and the directrices. Getting vertices from that should be easy. – ccorn Jan 20 '17 at 15:50
  • J. M., could you specify if the proofs for the axis slope and eccentricity equations involve vector calculus or linear algebra? I'd like to try deriving them myself, but I don't know either of those and don't want to waste time on it if it does require either of them. – harry Aug 22 '21 at 20:40

2 Answers2

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Appendix 1 of Keith Kendig's Conics seems to carry a lot of the formulae I want. On the other hand, I'd sure be interested in an even more extensive listing...

J. M. ain't a mathematician
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An old rich treatise is

  • John Casey: A treatise on the analytical geometry of the point, line, circle, and conic sections. Second edition 1893. Hodges, Figgis, & Co. (Ltd.) Dublin; Longmans, Green & Co. London; Dublin University Press.

That one also contains background for the complex tangent-through-focus topic which I find so useful in determining foci.

ccorn
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