Know your Surfaces

Completing the squares gives $(x+2y+z)^2+z^2-4(y+\frac{z}{2})^2=5$ . Since we have two positive and a negative coefficient, this is a hyperboloid of one sheet .

Background: Sylvester's Law of Inertia guarantees that any diagonal matrix congruent to the symmetric matrix $A$ of the quadratic form has the same number of positive, negative and zero diagonal entries... we need not go through the trouble of diagonalizing $A$ orthogonally (via eigenvalues).

We want to classify the signature of a quadratic form.

We have an equation of the form $x^T Ax = c$ , where $A$ is a symmetric real matrix and $c> 0$ . The classification will depend on the signs of the real eigenvalues of $A$ .

In this case, the matrix is $A = \begin{bmatrix}{1} && {2} && {2} \\ {2} && {0} && {0} \\ {1} && {0} && {1}\end{bmatrix}$ . By Cayley-Hamilton theorem,

$\det(A - \lambda I_3) = 0 \Rightarrow \begin{vmatrix}{1-\lambda} && {2} && {2} \\ {2} && {-\lambda} && {0} \\ {1} && {0} && {1-\lambda}\end{vmatrix} = 0\Rightarrow \lambda^3 - 2\lambda^2 - 4\lambda+ 4 = 0$

Let $f(\lambda) = \lambda^3 - 2\lambda^2 - 4\lambda+ 4$ . By trial and error, $f(-2) < 0 < f(-1), \quad f(0) > 0 > f(1), \quad f(2) < 0 < f(3) \; .$

Because $f(\lambda) = 0$ is cubic equation, then by bisection method , there exists exactly one root in each of the following intervals: $(-2,-1), (0,1), (2,3)$ . In other words, there are 3 eigenvalues: 1 of which is negative, 2 of which is positive. Thus the surface in question is a $\boxed{\text{Hyperboloid of 1 sheet}}$ .