# Homework Help: Does the discriminant test apply to this conic?

1. Jul 14, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Let's say I have 3x2 + 3y2 + 3z2 + 8xz - 8xy - 8zy

How can I apply the discirminant test to this 3d surface?

3. The attempt at a solution

In all of them, I keep getting 64 - 36 > 0

But there are like different kinds of hyperbola in 3d...

i don't want to use traces.

2. Jul 14, 2011

### vela

Staff Emeritus
That's not a conic section.

3. Jul 14, 2011

### flyingpig

Why?

4. Jul 15, 2011

### Dick

Why do you think it is? Conic sections are curves, not surfaces. You have a quadratic surface. That's not the same as a conic section.

5. Jul 15, 2011

### Staff: Mentor

flyingpig,
One thing that should be cleared up is what exactly are you working with.

This isn't an equation, so it's hard to say what you're working with.

If it's really 3x2 + 3y2 + 3z2 + 8xz - 8xy - 8zy = 0, then that is a surface in R3.

If it's w = 3x2 + 3y2 + 3z2 + 8xz - 8xy - 8zy, then the graph would require four dimensions.

6. Jul 15, 2011

### flyingpig

It is 3x2 + 3y2 + 3z2 + 8xz - 8xy - 8zy = 0, but isn't it just a conic in 3d...?

7. Jul 16, 2011

### Staff: Mentor

As Vela and Dick said, it's not a conic section, which is a curve. Here's a link to an article on Quadric Surfaces, one of which your equation represents.

Your equation represents a sphere, I'm pretty sure, but the xy, yz, and xz terms cause the thing to be tilted on all three axes.

8. Jul 16, 2011

### Char. Limit

Actually, checking on Wolfram-Alpha, it appears to be an elliptic cone.

9. Jul 16, 2011

### Staff: Mentor

You're going to have to convince me of this. The coefficients of all three squared terms are 1) positive and 2) equal, leading me to my assertion that the thing is a sphere. If it were a cone, I would expect the coefficient of one of the squared terms to be negative.

10. Jul 16, 2011

### I like Serena

My first instinct was that it should be a sphere as well.
However, there's this odd thing that the equation equals zero.....

So I solved the system and found the eigenvalues 11, -1, -1.
This means that we can reduce the system to the form:
11x2 - y2 - z2 = 0​
with an appropriate orthonormal base transformation (read: rotation).

This is not just an elliptic cone, it's a circular cone!

In retrospect I could have guessed, since only elliptic cones would contain (0,0,0).

Last edited: Jul 16, 2011
11. Jul 17, 2011

### flyingpig

How could sphere's axis be "tilted"?

12. Jul 17, 2011

### I like Serena

It can't!
It can only be translated.

So I think a quadric of a sphere can actually never have terms like xy, xz, or yz.
But it can have terms like x, y, and z.

Last edited: Jul 17, 2011
13. Jul 17, 2011

### HallsofIvy

Well, it can but since every diameter of a sphere is and "axis", it would still look exactly the same!

14. Jul 18, 2011

### nickalh

Arggghh, damn. The whole login again, lost my post.

I had a long explanation written agreeing with HallsOfIvy.

Also, it's hyperbolic.
See
solve for z and graph of 1 branch

Also, took me forever to find
the whole graph.

Doesn't that make it a variation of "hyperboloid of 1 sheet"?

15. Jul 18, 2011

### flyingpig

Wait so it is NOT a sphere now?!

16. Jul 18, 2011

### I like Serena

It is not a hyperboloid and it is not a sphere.

It is a circular cone (which is a degenerate hyperboloid).
Note that (0,0,0) is part of the surface.

17. Jul 19, 2011

### flyingpig

How do you check then? Without referring to another reference. My contour plots are rpetty ugly and bad...

18. Jul 19, 2011

### I like Serena

The method to classify a quadric surface is that you first bring the equation in the form:
$$\boldsymbol x Q \boldsymbol x^T + P \boldsymbol x^T + R = 0$$
I'm following wikipedia's notation here, where Q is a symmetric matrix, P is a column vector, and R is a scalar.

Then you determine the eigenvalues of Q.
Let's say they are $\lambda_1, \lambda_2, \lambda_3$.

The quadric will be a surface described by:
$$\lambda_1 x^2 + \lambda_2 y^2 + \lambda_3 z^2 + R' = 0$$
The difference is only a rotation and a translation.
I think R' is a different constant than the original R if you have a non-zero P.
(In your case R' is equal to R because P=0.)

From this you should be able to classify it yourself, or use e.g. wikipedia for a list of the cases.

There! I managed it without any references!
(Uhh, although I did refer to wikipedia. :uhh:)

Last edited: Jul 19, 2011
19. Jul 19, 2011

### flyingpig

Oh my god...

20. Jul 19, 2011

### I like Serena

Ah, I forgot, you were looking for a discriminant test.

Well, I found one here (hope you don't mind the reference):