Classifying quadric equations (when you have xy, zy bit etc)

Heya, sometimes I have trouble when I am given a quadric equation.. and i have to say what type of surface it is

Now I know the general forms of the quadric surfaces (ellipsoids, elliptic cones etc..). Their respective equations are usually of a form similar to (x/a)^2+(y/b)^2+(z/c)^2=1

So given an equation, I usually can convert it to required form by completing the square

But when an axz, byz, cyz (a,b, c constants) are thrown in I have no idea what to do.. I cannot complete the square and get it into any recognisable form

To see what i mean, in the picture attached, number 2 and 3 are fine, but 1 and 4 leave me clueless

Any help would be great?

I do not know what is your current/previous knowledge but the only way I know is by removing the cross terms by diagonalizing the associated 3x3 matrix (finding eigenvalues and eigenvectors).

it practically impossible to run you through this method here

I suggest

google "diagonilising quadratic"
look at a "glossy type american book" on "linear algebra.

ok thanks, I had a feeling it had something to do with matrices

I am aware about the associated 3 by 3 matrix..... then by getting it into diagonal form i can find associated equations which should be a simpler equation or something?

thanks

yes

it can be writen as (X,Y,Z)^TD(X,Y,Z)

where (X,Y,Z) denotes new coordinates
D is a diagonal matrix

(these types are usually rotated quadrics)

Ok cheers

And say if we had quadratic equation of 3 variables (x,y and z) with the constant term k

Could we just ignore it the k for now, find diagonal of 3 by 3 matrix then add it back on?
Cause otherwise you would get a complicated 4 by 4 matrix

what do you mean with the constant term k?

type an example

x^2+y2+z2+4xz+k=0

I was thinking the matrix would be xT A x
where x=(x,y,z,1)

Or could we ignore the constant bit, diagonalise with x=(x,y,z) then add constant back on at the end

Not sure if im making sense

Still a 3x3

look at google documents or if you have access to a book for an example

it is not that hard when you see a fully worked example

Diagonalizing will end up with you being able to rewrite the equations in a form

$\alpha ({\bf x.a})^2 +\beta ({\bf x.b})^2 + \gamma ({\bf x.c})^2 = C$ for suitable constant vectors a, b, c and constant scalars $\alpha, \beta, \gamma, C$.

(which is basically the same as $\alpha x^2 + \beta y^2 + \gamma z^2 = C$ but with a change of basis).

thanks guys I think i got it