Can You Classify Quadrics Based on Second Order Equations?

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Discussion Overview

The discussion revolves around the classification of quadrics based on general second order equations in three variables. Participants explore methods for transforming these equations into canonical forms, such as ellipsoids and hyperboloids, through translations and rotations.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant seeks resources for classifying quadrics from second order equations.
  • Several participants provide links to external resources related to conics and quadrics.
  • A participant clarifies the need for transforming a general second order equation in three variables into canonical forms by removing mixed terms.
  • Another participant suggests that the process involves completing the square in quadratic equations, providing a detailed explanation for both one and two variables.
  • A later reply indicates that for three variables, diagonalizing the matrix representing the quadric is necessary, and computing the eigenvalues is part of this process.

Areas of Agreement / Disagreement

Participants generally agree on the need to diagonalize the matrix for three variables, but there are varying levels of detail and understanding regarding the methods for classification and transformation.

Contextual Notes

The discussion includes assumptions about familiarity with quadratic equations and matrix operations, which may not be explicitly stated. There are also unresolved steps in the transformation process for three variables.

Belgium 12
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Hi,

can someone help me on a website for the classification of quadrics(ellipsoids,paraboloids,...) starting from the general second order equation.

Thanks
 
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Hi,

What I really mean is:

you have a general equation of the second order in three variables and by translations

and rotations to remove Xy-Xz and yz terms you transform it to the canonical forms

of quadrics.(ellipsoids,hyperboloids,cone...)

Bye
 
do you want to know how to complete the square in a quadratic equation? that's all it is.
 
in one variable you start from X^2 + bX + c, and you get

X^2 + bX + (b/2)^2 +c - (b/2)^2 = (X+ b/2)^2 + [(4c-b^2)]/4.

in 2 variables you start from X^2 + bXY + cY^2, and get

X^2 + bXY +(bY/2)^2 + cY^2 - (bY/2)^2

= (X+bY/2)^2 + [4c-b^2]/4 Y^2.

so now you replace X+ bY/2 by W and you have W^2 + [4c-b^2]/4 Y^2.

there are no WY terms, and so you can tell which quadric you have by the sign of

[4c-b^2]/4.

maybe you want to expand this a little, by throwing in a third letter Z, but this is the main trick.

i guess for a general homogeneous function of three variables you want to diagonalize the matrix representing the quadric.

the previous discussion above works for Z = quadratic in (X,Y), which arises in calculus of two variables.
 
Last edited:
Hi,
mathwonk

yes that's.for three variables I think you have to diagonalize the matrix.First you you must

compute the eigen-values of the matrix .

Thank you for your help.
 

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