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Can anybody fully describe the three dimensional surface represented by the equation $a/x+b/y+c/z=0$ where $a,b,c$ are real numbers?

Shahul Hameed
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    Can you please show some of your own working here. How have you tried to solve this problem? – EHH Nov 22 '17 at 15:18
  • Even if it is not a classical surface how the surface will look like for different values of fixed constants a, b and c? When I tried with online Wolfram software they have given some examples and tells that it represents an infinite elliptic cone. I am not sure about it. – Shahul Hameed Nov 22 '17 at 15:23
  • @JeanMarie I put an answer, which should be enough if the OP is willing to put in some work himself – Will Jagy Nov 22 '17 at 16:47
  • A good way to analyze a surface given by an implicit equation as is the case here is to take slices, by assigning a value to one of the variables, for example here take $z=1$ ; in this way, you have to tackle only two variables. – Jean Marie Nov 22 '17 at 17:05

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With the origin and the axes removed, it is the cone $$ ayz + b zx + c xy = 0. $$ Note that, if two out of three of $x,y,z$ are zero, then the form is zero. For example, if $x,y$ are both zero, we are on the $z$ axis somewhere. This point is in the cone, it is just not allowed in the original description of the surface, which prohibited any coordinate being zero.

The Hessian matrix of the quadratic form is $$ H = \left( \begin{array}{ccc} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{array} \right) $$ For unspecified values of $a,b,c$ the eigenvalues are unpleasant, $$ \lambda^3 - \left(a^2 + b^2 + c^2 \right) \lambda - 2abc = 0. $$ It is possible, with integers $a,b,c,$ to get at least one rational eigenvalue; when $a=b,$ the eigenvalues are $(-c, \frac{c \pm \sqrt{8a^2 + c^2}}{2})$ In general, or using the symbols, this is Casus Irreducibilis. Hmmm. This point seems worth further investigation...

On the other hand, for chosen values of $a,b,c,$ things sometimes come out nicely. For example, with $a=b=c=1,$ we have $$ H = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \right) $$ and the eigenvalues are $-1,-1,2.$ In this case, as rows, the eigenvectors (normalized) can be taken as $(1,1,1) / \sqrt 3, \; \; $ $(-1,1,0) / \sqrt 2, \; \; $ $(-1,-1,2) / \sqrt 6. \; \; $

If we were setting the quadratic form to a nonzero constant, we would need to know the signs of the eigenvalues, as in Sylvester's Law of Inertia, to say whether we had a hyperboloid of one or two sheets. Since we are setting to zero, this does not matter: it is just the cone, although the origin and axes are removed because the original problem did not allow the variables to be zero.

Will Jagy
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