Classification Of $3x^2+4xy-6y^2-z=0$

Question

Classify the quadratic surface $3x^2+4xy-6y^2-z=0$

Looking at $$\begin{vmatrix}\lambda-3 & -2 & 0\\-2 & \lambda+6 & 0\\ 0 & 0 & 0\end{vmatrix}=0$$

So how can I classify it?

Answer 1

First, as the matrix $$\begin{bmatrix}3&0&0&0\\2&-6&0&0\\0&0&0&\frac12\\0&0&\frac12&0\end{bmatrix}$$ has rank $4$, the quadric is non-degenerate.

Second, as the characteristic polynomial of the quadratic form is $_\lambda(\lambda^2+3\lambda-22)$, the eigenvalues are $0$ and a positive and a negative eigenvalue (just look at the signs of the coefficients). The signature of the quadratic form is $(1,1)$ and its rank is $2$.

So we know the quadric is either a hyperbolic paraboloid or a hyperbolic cylinder.

The former is correct, because the quadric has an equation of the form $$z=\text{quadratic form in } x \text{ and } y \text{ alone.}$$

Indeed its equation can be written as $$z=\frac12(3x+2y)^2-\frac{22}3y^2.$$