Talk:Quadric

message: "Geometrically, every non-degenerate quadric surface is obtained from a conic section in the plane as a surface of revolution, followed by an affine transformation."
I am skeptical.  What are the conic section and affine transformation that produces a hyperbolic parabaloid?
192.249.47.195 (talk) 18:46, 28 September 2009 (UTC)
thread_title:  Hyperbolic Parabaloid and Affine Transformations 

message: The colors in the image in the Euclidean Plane section are mismatched from the description of the image. 209.6.73.123 (talk) 22:49, 4 September 2018 (UTC)
replies: text: It is fixed now.  Thank you for pointing this out.—Anita5192 (talk) 23:28, 4 September 2018 (UTC)
replies: text:  Perhaps the color "goldenrod" is better: \color{goldenrod}{abc}. Please try or look at Help:Displaying a formula.  --Ag2gaeh (talk) 07:24, 5 September 2018 (UTC)
replies: text: Goldenrod is less difficult to see, but still difficult.  Actually I don’t think the image needs a color-coordinated caption.  If the colors are completely removed from the caption, the specific eccentricities clarify the correspondence.  I suggest we make the caption text entirely black.—Anita5192 (talk) 16:05, 5 September 2018 (UTC)
thread_title:  Color Mismatch 

message: I tried to improve section Projective quadrics over fields. If there is no objection, I shall remove the tag on "multiple issues".--Ag2gaeh (talk) 09:56, 2 March 2019 (UTC)
thread_title:  Revision of section "Projective quadrics over fields" 

message: In the current version of the article I read the sentence: "This define affine quadrics." Either grammatically or spelling-wise there seems to be an error in this sentence, which obscures its meaning to me. What did the author intend to write?
Was it "This defines affine quadrics."?
Or was it "These define affine quadrics."?
Or anything else?Redav (talk) 09:14, 26 April 2020 (UTC)
replies: text: The first interpretation is the correct one. However, as "This" is ambiguous, I have moved the definition at the beginning of the section, and added explanations for clarifying the end of the section. D.Lazard (talk) 09:48, 26 April 2020 (UTC)
thread_title:  Grammar and/or Spelling 

message: has recently added the following paragraph in section :
I'll remove this paragraph because of the following issues, which, all together, make that this paragraph does not belong to this section.
* The section is about rational points of quadrics, and the primitive Heronian triangles are in one to one correspondence with the rational points of the quartic obtained by squaring Heron's formula. As a quartic is not a quadric, the relationship between Heronian triangles and rational points (the subject of this section) does not belong to this article.
* The assertions contained in the paragraph are true, but not evident (I had to think about a while for finding a proof). So they must be sourced, proved, linked, or easy to find in Heron triangle. This is not the case, and the paragraph does follows the policy of WP:Verifiability. 
* It is not said that the relation x_1 x_2 + x_2 x_3 + x_1 x_3 = 1 is satisfied by half-angle tangents of every triangle.
* Parametrizing this quadric does not requires the method described in this article, as the parametrization x_1=t, x_2=u, x_3=\frac{uv}{u+v} is evident.
* Several rational parametrizations of the "Heronian quartic" are given in Heronian triangle, but it is unclear whether any of them results from a paramatrization of the quadric x_1 x_2 + x_2 x_3 + x_1 x_3 = 1. So this way of parametrizing heronian triangles seems WP:OR. 
D.Lazard (talk) 15:49, 2 December 2022 (UTC)
replies: text: I agree that if it isn't WP:CALC then an editor can request that there be a citation.  Go ahead and remove it if you wish.  It's been a while since I've had my algebraic varieties class, so odds are long that I will find a Wikipedia-quality citation, but I'll keep my eyes open.  (Yes, we could clarify that x_1 x_2 + x_2 x_3 + x_1 x_3 = 1 is satisfied by every triangle; it leads to a Heronian triangles if and only if each of the variables is positive and rational.  Yes, there is a way to frame Heronian triangles as using a quartic equation, but that doesn't make the quadratic approach any less valuable.  Yes, there are other ways to solve this quadratic, but there are other ways to solve the circle too; e.g., x_1 = u, x_2 = \pm\sqrt{1-u^2}) —Quantling (talk | contribs) 16:21, 2 December 2022 (UTC)
thread_title:  Heronian triangles