what is the surface representing $a/x+b/y+c/z=0$?

Question

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?

Answer 1

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.