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  1. A necessary definition

$F(x,y,z)$ is called a homogeneous function of degree $n$ ($n$ is an integer) if $F(tx,ty,tz)=t^nF(x,y,z)$ holds for any nonzero real number t,and any possible x , y, z. In this case, $F(x,y,z)=0$ is said to be a homogeneous equation. For example, $x^2+y^2+z^2=F(x,y,z)$ is a homogeneous function of degree $2$.

  1. The question.

"In a rectangular coordinate system with the vertex of the conical surface as the origin,the conical surface can be represented by homogeneous equations of $x$, $y$, and $z$". This is a conclusion we learnt from our textbook on Analytic Geometry, which appears as a theorem. However, we have no idea why it's correct, whether this theory only holds in rectangular coordinate system...we can't find any proof of it in any literature,and the writer of the textbook didn't provide proof as well.

  1. Attempts already made

According to the rotational surface equation "$kz=±\sqrt{(x'^2+y'^2)}$" given in the rectangular coordinate system, it can be proved easily that this conclusion holds for the circular conical surface (revolution conical surface) in the rectangular coordinate system, but for other (general) conical surfaces I can’t think of a good proof method.

So if anyone understands it and knows why, please answer my question (better off if you can provide me with a rigorous thinking or proof:)) Thank you soooooo much!!

FishDrowned
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Lily Luo
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  • so, you haven't had linear algebra yet? – Will Jagy Nov 18 '23 at 05:12
  • @willj Sorry, I don't know what you meant. 'Have I learned linear algebra' or 'Have I tried to prove it with the relevant knowledge of linear algebra'? If it is the former, I learned linear algebra in Advanced Algebra; if it is the latter, I may not be able to accurately say whether I have used the knowledge of linear algebra, but indeed I did not try to prove it with the knowledge of matrix or such similar knowledge. – Lily Luo Nov 19 '23 at 01:16
  • alright, an indefinite ternary quadratic form, such as $x^2 - 3 y^2 + 7 z^2,$ when non-degenerate, has a cone as its zero set. The Hessian matrix of second partial derivatives is just a bunch of constant coefficients; nondegenerate means this matrix, call it $H,$ is nonsingular. Your quadratic form is then $(1/2) X^T HX,$ where this $X$ is the column vector with entries $x,y,z.$ Your $kz$ thing is just $x^2 + y^2 - k^2 z^2$ Oh, change of variable is $H \mapsto P^T HP$ where $P$ is nonsingular... and on and on – Will Jagy Nov 19 '23 at 01:29
  • As with the example $x^2 - 3 y^2 + 7 z^2$ I gave, a quadratic form usually gives a cone with elliptical cross section. – Will Jagy Nov 19 '23 at 01:31
  • @willj Thanks for your comments! I'll try again with your method. – Lily Luo Nov 19 '23 at 01:51
  • @willj Just one question for your last sentence "a quadratic form usually gives a cone with elliptical cross section". As we all know, the directrix of a cone can be a fixed curve of any shape, if it is a strange shape rather than a circular or elliptical shape (such as the shape of a cloud?), it may not be able to cut out elliptical shapes, right? – Lily Luo Nov 19 '23 at 02:10
  • What textbook? I still don't really know what you mean by a cone – Will Jagy Nov 19 '23 at 02:16
  • oh, well. I think you would profit from some of these books, the easiest of the three is Linear Geometry by Artzy, a review.. https://maa.org/press/maa-reviews/linear-geometry . I believe they are talking about the null set of a quadratic form (indefinite) because they want – Will Jagy Nov 19 '23 at 02:33
  • @willj "A curved surface produced by a family of straight lines in space passing through a fixed point (vertex) and intersecting a fixed curve (directrix) is called a cone". This is the definition from our textbook Analytic Geometry written by a Chinese professor Qiu Weisheng. （And I'm now warned not to extend discussions in comments. So thank you again first, and I think or worry about maybe I will be banned from replying at any time.） – Lily Luo Nov 19 '23 at 02:36
  • well, the condition that "the conical surface can be represented by homogeneous equation" is rather stronger than just being a bunch of lines through a vertex and some wavy curve. – Will Jagy Nov 19 '23 at 02:51

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