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

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    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?
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    (Original post by number23)
    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.
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    (Original post by TeeEm)
    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
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    (Original post by number23)
    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)
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    (Original post by number23)
    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?
    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).
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    (Original post by TeeEm)
    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
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    (Original post by number23)
    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
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    (Original post by TeeEm)
    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
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    (Original post by number23)
    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
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    (Original post by TeeEm)
    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

    (Original post by DFranklin)
    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
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    (Original post by number23)
    thanks guys I think i got it
    no worries
 
 
 
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