Talk:Conic section

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{{edit semi-protected}} Please add the reference to this MATLAB Central URL containing the code to detect conics intersection:

http://www.mathworks.com/matlabcentral/fileexchange/28318-conics-intersection

Pierluigi (talk) 8:52, 29 November 2010 (UTC)

It did confuse me, anyway. It starts with :

> In homogeneous coordinates a conic section can be represented as:

 ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dxz+Eyz+Fz^{2}=0.}$

But that is the equation of a surface, not a curve. In fact, if I'm not mistaken it's the equation of the cone of whom the conic is a section with a plan. It's easy to see when we notice that the matrix is symmetric and thus can be diagonalized by an orthogonal matrix, with real eigenvalues. Necessarily at least one eigenvalue is negative (otherwise we have a sphere of nul radius, that is just a point), and we have the equation of a cone.

I think this ought to be clarified. Grondilu (talk) 05:37, 22 December 2022 (UTC)Reply

You have to know what homogeneous coordinates and homogeneous polynomials are first. We have the implicit curve:

${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$

But this has 1 term of degree 0, 2 terms of degree 1, and 3 terms of degree 2.

By adding new variables ${\displaystyle x,y,z}$ and replacing ${\displaystyle x\to x/z}$ and ${\displaystyle y\to y/z}$ (after this replacement the ${\displaystyle x,y}$ here now stand for something slightly different than the originals), we get:

${\displaystyle Ax^{2}z^{-2}+Bxyz^{-2}+Cy^{2}z^{-2}+Dxz^{-1}+Eyz^{-1}+F=0.}$

Then by multiplying everything by ${\displaystyle z^{2},}$ we can make the polynomial on the left hand side homogeneous (every term has degree 2):

${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dxz+Eyz+Fz^{2}=0.}$

This is still intended to represent the same implicit curve as the original. We are just using a different coordinate system. –jacobolus (t) 05:58, 22 December 2022 (UTC)Reply

You're right. I guess I forgot that in homogeneous coordinates, there is one additional dimension so the equation looks like it's one dimension larger (a surface instead of a curve, in that case). I suppose the section as it is now is fine, then.--Grondilu (talk) 11:17, 22 December 2022 (UTC)Reply

Maybe someone who is an expert (not me) can still try to clarify and elaborate, explaining why we want the polynomial to be homogeneous and what else we can do with it. –jacobolus (t) 17:37, 22 December 2022 (UTC)Reply

In the section "euclidean standard forms". Should be "y^2 = 4ax", not "y = 4ax"

187.116.67.44 (talk) 00:04, 14 August 2024 (UTC)Reply

Thanks, that was my typo, when I made a higher-resolution version of the image. Fixed. –jacobolus (t) 01:41, 14 August 2024 (UTC)Reply

I don't know how this works but this page is not editable for me, so I have to say here that somewhere in there the complex plane is called ℂ² instead of ℂ 24.56.238.67 (talk) 00:38, 1 December 2024 (UTC)Reply

The sentence was correct, although confusing. Indeed, "complex plane" has several different meanings in mathematics: it may refer to the complex numbers viewed as a Euclidean plane over the reals. It may also refer to an affine or projective plane over the complexes. Here, ⁠${\displaystyle \mathbb {C} ^{2}}$⁠ is a specific complex affine plane, often called complex coordinate plane. Therefore, I changed "complex plane" into "complex coordinate plane". D.Lazard (talk) 11:11, 1 December 2024 (UTC)Reply

If the angle between the surface of the cone and its axis is alpha and the angle between the cutting plane and the axis is beta the eccentricity is cos(alpha)/cos(beta) that's certainly wrong! because for an ellipse we have alpha<beta then cos(alpha)>cos(beta) then we get that the eccentricity is larger than 1! which is incorrect for an ellipse. I tried proving this and I found that the eccentricity is equal to sin(alpha)/sin(beta) Issam Zreik (talk) 14:45, 10 June 2025 (UTC)Reply

Here's the source, which looks reasonable at a glance. If ⁠${\displaystyle \alpha }$⁠ is the angle between the cutting plane and the axis and ⁠${\displaystyle \beta }$⁠ is the angle between the cone and the axis, then for an ellipse ⁠${\displaystyle \alpha >\beta }$⁠. –jacobolus (t) 23:41, 10 June 2025 (UTC)Reply

yes, I understood. I made a mistake in the proof. I apologize and thank you for your time. Issam Zreik (talk) 03:04, 11 June 2025 (UTC)Reply