I have no idea what below surface equation represent

Question

I have the equation:

$$x^2+y^2+4z^2-14xy+8xz-8yz=24$$

What does this equation represent?

How can I find the "axes" of it (?), and is it possible to draw it when it intersect the plane $z=0$?

Answer 1

Hint: Complete the square.

As a short example $$x^2 - 14xy = (x - 7y)^2 - 49y^2$$

Answer 2

This is the equation of a quadric surface. It corresponds to a quadratic form $q(x,y,z)$ which can be represented by the symmetric matrix: $$A=\begin{bmatrix}1&-7&4\\-7&1&-4\\4&-4&4\end{bmatrix}$$ To find the axes of the quadric, you have to write its equation in normalised form, as sum of squares of linear form (Gauß's reduction): if $$q(x,y,z)=\lambda_1 \ell_1^2(x,y,z)+\lambda_2 \ell_2^2(x,y,z)+\lambda_3 \ell_3^2(x,y,z)$$ is the reduced form (here we obtain the sum of the squares of $3$ forms), the equations of the axes are simply $\ell_i^2(x,y,z)=\ell_j^2(x,y,z)=0$ for all pairs $i\neq j$. The type of quadric surface is determined by the signature of the quadratic form, wwhich the pair $(p,q)$, where $p$ is the number of squares of linear forms with a positive coefficient, and $q$ the number of forms with a negative coefficient. It is also the number of positive and negative eigenvalues.

Here, the characteristic polynomial is $-x^3+6x^2+72x\,$ and the eigenvalues are $12,-6,0$. The eigenvectors are, respectively: $$\begin{bmatrix}1\\-1\\1\end{bmatrix},\quad\begin{bmatrix}1\\1\\0\end{bmatrix},\quad\begin{bmatrix}1\\-1\\-2\end{bmatrix}$$

In this (orthogonal) basis the matrix of the quadratic form is $$\begin{bmatrix}12&0&0\\0&-6&0\\0&0&0\end{bmatrix}$$ hence the equation of the surface is $\,36X^2-6Y^2=24\iff \dfrac32X^2-\dfrac14Y^2=1$, which is the equation of a hyperbolic cylinder.