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Does the discriminant test apply to this conic?

  1. Jul 14, 2011 #1
    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. jcsd
  3. Jul 14, 2011 #2

    vela

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    That's not a conic section.
     
  4. Jul 14, 2011 #3
    Why?
     
  5. Jul 15, 2011 #4

    Dick

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    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.
     
  6. Jul 15, 2011 #5

    Mark44

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    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.
     
  7. Jul 15, 2011 #6
    It is 3x2 + 3y2 + 3z2 + 8xz - 8xy - 8zy = 0, but isn't it just a conic in 3d...?
     
  8. Jul 16, 2011 #7

    Mark44

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    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.
     
  9. Jul 16, 2011 #8

    Char. Limit

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    Actually, checking on Wolfram-Alpha, it appears to be an elliptic cone.
     
  10. Jul 16, 2011 #9

    Mark44

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    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.
     
  11. Jul 16, 2011 #10

    I like Serena

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    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). :smile:
     
    Last edited: Jul 16, 2011
  12. Jul 17, 2011 #11
    How could sphere's axis be "tilted"?
     
  13. Jul 17, 2011 #12

    I like Serena

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    It can't! :smile:
    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
  14. Jul 17, 2011 #13

    HallsofIvy

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    Well, it can but since every diameter of a sphere is and "axis", it would still look exactly the same!
     
  15. Jul 18, 2011 #14
    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.
    In the future, please always post your wolfram links!!

    Doesn't that make it a variation of "hyperboloid of 1 sheet"?
     
  16. Jul 18, 2011 #15
    Wait so it is NOT a sphere now?!
     
  17. Jul 18, 2011 #16

    I like Serena

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    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.
     
  18. Jul 19, 2011 #17
    How do you check then? Without referring to another reference. My contour plots are rpetty ugly and bad...
     
  19. Jul 19, 2011 #18

    I like Serena

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    The method to classify a quadric surface is that you first bring the equation in the form:
    [tex]\boldsymbol x Q \boldsymbol x^T + P \boldsymbol x^T + R = 0[/tex]
    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 [itex]\lambda_1, \lambda_2, \lambda_3[/itex].

    The quadric will be a surface described by:
    [tex]\lambda_1 x^2 + \lambda_2 y^2 + \lambda_3 z^2 + R' = 0[/tex]
    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! :smile:
    (Uhh, although I did refer to wikipedia. :uhh:)
     
    Last edited: Jul 19, 2011
  20. Jul 19, 2011 #19
    Oh my god...
     
  21. Jul 19, 2011 #20

    I like Serena

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

    Well, I found one here (hope you don't mind the reference):
    http://mathworld.wolfram.com/QuadraticSurface.html

    I'm afraid it may not be as simple as you'd like though. :wink:
    In effect it's based on the method I just described.
     
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