"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,but we have no idea why it is right,why it can be treated as a theorem,whether this theory only holds in rectangular coordinate system...My geometry teacher said that he can't find any proof of this "theorem" in any literature,and the writer of the textbook didn't provide proof as well.
So if anyone understands it and knows why,please answer my question and provide me with a rigorous proof:)Thank you sooooooooooooo much!!!!!!!!
Ok, now I realised that maybe I should add something necessary to make the question be more concrete and clear.
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)=tⁿF(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, x²+y²+z²=F(x,y,z) is a homogeneous function of degree 2.
2.The definition of 'conical surface'(or 'cone') in the textbook
"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."
3.Attempts already made
According to the rotational surface equation "kz=±√(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.
Peter Breuer Firstly thanks for you reply!!
I can roughly understand what you said. As for your confusion about the definition of "cone", actually I don't know how to explain it clearly, because that's just how it is defined in our textbook (btw I know that the definition you mentioned is more commen ). And as a result of its generality, a general "cone" is actually unable to write a general formula like 'F(x) = (x x u)2-k2(x.u)2 = 0', which is the biggest difficulty in proving.
Meanwhile I asked the writer of the textbook, and he gave me the reply as the following:
"The surfaces studied in this book are algebraic surfaces, that is, the left end of the surface equation is a polynomial. In the rectangular coordinate system with the vertex of the cone as the origin, let the left end of the equation F (x, y, z) = 0 of the cone surface S be a polynomial of degree n, and use the plane z = h (h > 0) to cut the cone surface S to obtain the curve C.Taking this curve C as the directrix of the cone surface S, the equation of the cone surface S is solved, and then it is proved that this equation is a homogeneous equation of degree n."
Peter Breuer
I‘ll try to understand what you are saying, because it literally involves something beyond what I have learned. (Sorry for my lack of academic knowledge. T^T) And there is one more little question:
I discussed your initial answer with my teacher, he thought maybe you initially mistakenly placed the question in the context of non-Euclidean geometry (but as a matter of fact we are talking about Euclidean geometry). So I want to ask you whether my teacher's idea about your first answer is right? Because after all this theorem is obviously true in non-Euclidean geometry:D
Finally thank you again for your continuously serious answers!