Talk:Conic section

Edit request to Conics Intersection paragraph, 29 November 2010

{{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)

Type of conic for given equation ax^2+bxy+cy^2+dx+ey+f=0

For ax^2+bxy+cy^2+dx+ey+f=0:

Let D=b^2-4ac and Q=det(a,b/2,d/2;b/2,c,e/2;d/2,e/2,f) and R=d^2+e^2-4(a+c)f

1.

D>0 --> goto 2 D=0 --> goto 3 D<0 --> goto 4

2.

Q=0 --> (two intersecting lines) Q≠0 --> (hyperbola)

3.

Q=0 --> goto 5 Q≠0 --> (parabola)

4.

Q=0 --> (a single point) Q≠0 --> goto 6

5.

R>0 --> (two parallel straight lines) R=0 --> (a single line) R<0 --> (empty set)

6.

(a+c)Q>0 --> (empty set) (a+c)Q<0 --> goto 7

7.

a=c and b=0 --> (circle) a≠c and/or b≠0 --> (ellipse)

Is it right? 2402:7500:92E:A23D:349B:A9A7:2AAF:1C1E (talk) 12:53, 11 September 2021 (UTC)

This is hardly human readable, and seems almost correct (the case $a=b=c=0$ is ommitted). In any case this classification is given in § Discriminant for the case $Q\neq 0$ (true conics), with a link to Degenerate conic § Discriminant for the degenerate cases. Maybe, the complete classification should be made more visible, but your style is definitively not convenient for Wikipedia. D.Lazard (talk) 15:30, 11 September 2021 (UTC)

Some psychological tests use the similar thing, the correct thing should be:

For ax^2+bxy+cy^2+dx+ey+f=0:

Let D=b^2-4ac and Q=det(a,b/2,d/2;b/2,c,e/2;d/2,e/2,f) and R=d^2+e^2-4(a+c)f

1.

D>0 --> goto 2

D=0 --> goto 3

D<0 --> goto 4

2.

Q=0 --> (two intersecting straight lines)

Q≠0 --> (hyperbola)

3.

Q=0 --> goto 5

Q≠0 --> (parabola)

4.

Q=0 --> (a single point)

Q≠0 --> goto 6

5.

a=0 and b=0 and c=0 --> goto 7

a≠0 and/or b≠0 and/or c≠0 --> goto 8

6.

(a+c)Q>0 --> (empty set)

(a+c)Q<0 --> goto 9

7.

d=0 and e=0 --> goto 10

d≠0 and/or e≠0 --> (a single straight line)

8.

R>0 --> (two parallel straight lines)

R=0 --> (a single straight line)

R<0 --> (empty set)

9.

a=c and b=0 --> (circle)

a≠c and/or b≠0 --> (ellipse)

10.

f=0 --> (the full plane)

f≠0 --> (empty set)

Note: In step 6, (a+c)Q cannot be =0, since D<0, and D<0 implies ac>0, thus a+c cannot be =0 since a and c are real numbers, also Q is not 0

114.41.119.57 (talk) 09:13, 12 September 2021 (UTC)

Wikipedia is not a psychological test. It requires prose. D.Lazard (talk) 09:47, 12 September 2021 (UTC)

I've (range-)blocked both of the IPs involved here for block evasion, see Wikipedia:Sockpuppet investigations/Xayahrainie43. --Blablubbs (talk) 12:27, 12 September 2021 (UTC)

Why does "semilatus rectum" redirect here?

What does it mean? Term not used in article. Equinox ◑ 23:15, 24 October 2022 (UTC)

To quote the article, "The latus rectum is the chord parallel to the directrix and passing through a focus; its half-length is the semi-latus rectum (ℓ)." –jacobolus (t) 00:24, 25 October 2022 (UTC)

Semi-protected edit request on 25 November 2022

change "ooooooo" to "usually" in the section "intersection at infinity" Atobi16 (talk) 10:57, 25 November 2022 (UTC)

"oooooooo" removed. D.Lazard (talk) 12:37, 25 November 2022 (UTC)

The section about homogeneous coordinates may be confusing

It did confuse me, anyway. It starts with :

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

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

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)

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

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

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

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

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

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

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

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)

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)

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)