Kuzee
Vector Algebra and Analytic Geometry of Space

Methods of Classification

Reducing to the Standard Form

Coordinate Transformations

Through an orthogonal transformation, the equation of a general quadric surface can be reduced to its standard form:

The general equation of a quadric surface

Ax2+By2+Cz2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0

Through a coordinate transformation, the cross terms can be eliminated and the equation reduced to:

The standard form

λ1x′2+λ2y′2+λ3z′2+λ4x′+λ5y′+λ6z′+λ7=0\lambda_1 x'^2 + \lambda_2 y'^2 + \lambda_3 z'^2 + \lambda_4 x' + \lambda_5 y' + \lambda_6 z' + \lambda_7 = 0

The Eigenvalue Test

The classification theorem for quadric surfaces

The type of a quadric surface is determined by the eigenvalues λ1,λ2,λ3\lambda_1, \lambda_2, \lambda_3 of its quadratic form.

The Classification Rules

According to the signs and the counts of the eigenvalues:

  1. The ellipsoid: the three eigenvalues are of the same sign (J=λ1λ2λ3>0J = \lambda_1\lambda_2\lambda_3 > 0)

    • a real ellipsoid: the equation has real solutions
    • an imaginary ellipsoid: the equation has no real solutions

  2. The hyperboloid: the three eigenvalues are two of one sign and one of another (J<0J < 0), with a nonzero constant term

    • the hyperboloid of one sheet / of two sheets: the two eigenvalues of the same sign are either two positive and one negative or one positive and two negative, and the sign of the constant term distinguishes the two

  3. The cone: the three eigenvalues are two of one sign and one of another (J<0J < 0), and the constant term is zero

    • the elliptic cone (passing through the origin)

  4. The paraboloid: exactly one eigenvalue is zero (J=0J = 0)

    • the elliptic paraboloid: the other two eigenvalues are of the same sign
    • the hyperbolic paraboloid (the saddle surface): the other two eigenvalues are of different signs

  5. The cylinder: a degenerate surface (at least one eigenvalue is zero, and there is no linear term in the corresponding direction)

    • the elliptic cylinder: one eigenvalue is zero and the other two are of the same sign
    • the hyperbolic cylinder: one eigenvalue is zero and the other two are of different signs
    • the parabolic cylinder: two eigenvalues are zero and one is nonzero, and there is a linear term

The Invariant Test

The Matrix of the Quadratic Form

For the equation:

Ax2+By2+Cz2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0

the matrix of the quadratic form is:

(AD/2E/2D/2BF/2E/2F/2C)\begin{pmatrix} A & D/2 & E/2 \\ D/2 & B & F/2 \\ E/2 & F/2 & C \end{pmatrix}

The Invariants

The type of the surface is decided by computing invariants of the matrix such as its trace and its determinant.

The Method of Direct Observation

Analysing the Form of the Equation

Judge directly by looking at the form of the equation:

  • the sphere: the coefficients of x2,y2,z2x^2, y^2, z^2 are equal
  • the ellipsoid: the coefficients of the squares of all three variables are positive
  • the hyperboloid of one sheet: the coefficients of the squares of two variables are positive and one is negative
  • the hyperboloid of two sheets: the coefficients of the squares of two variables are positive and one is negative, and the constant term is −1-1
  • the paraboloid: it contains a linear term in one of the variables
  • the cylinder: the equation is missing one of the variables
  • the cone: the constant term is zero and the coefficient of one of the variables is negative

Exercises

Exercise 1

Determine the type of the surface 2x2+3y2+4z2−4xy+2xz−6yz+8x−10y+12z+20=02x^2 + 3y^2 + 4z^2 - 4xy + 2xz - 6yz + 8x - 10y + 12z + 20 = 0.

quadric surfacethe methods of classificationthe general equation

Approach:

The general equation has to be reduced to its standard form, and then the type judged from the eigenvalues.

Steps:

  1. This is the general equation of a quadric surface, and it has to be reduced to its standard form by a coordinate transformation
  2. Compute the eigenvalues of the matrix of the quadratic form
  3. Judge the type of the surface from the signs of the eigenvalues

Answer: the type can only be determined after the concrete computation.

Exercise 2

Explain why the coefficients of x2,y2,z2x^2, y^2, z^2 in the equation of a sphere must be equal.

quadric surfacethe spherethe relations between the coefficients

Approach:

Derive it from the geometric definition of the sphere and its standard equation.

Steps:

  1. A sphere is the set of the points whose distance from a fixed point equals a fixed length
  2. The distance formula: (x−a)2+(y−b)2+(z−c)2=r\sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2} = r
  3. Squaring: (x−a)2+(y−b)2+(z−c)2=r2(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2
  4. Expanding: x2+y2+z2−2ax−2by−2cz+(a2+b2+c2−r2)=0x^2 + y^2 + z^2 - 2ax - 2by - 2cz + (a^2+b^2+c^2-r^2) = 0
  5. Hence the coefficients of x2,y2,z2x^2, y^2, z^2 must be equal

Answer: because a sphere is isotropic and symmetric about the three coordinate axes, so the three squared coefficients must be equal

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
λ1,λ2,λ3\lambda_1, \lambda_2, \lambda_3EigenvalueseigenvaluesThe eigenvalues of the matrix of the quadratic form
Ax2+By2+Cz2+⋯Ax^2 + By^2 + Cz^2 + \cdotsGeneral equationgeneral equationThe general form of the equation of a quadric surface
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