Restricting Coefficients in a General Second Degree Equation

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The discussion focuses on the general second-degree equation Ax^2 + Bxy + Cy^2 = 0, specifically exploring restrictions on coefficients A, B, and C. It is established that any values of A, B, and C will yield a set of points in the xy-plane satisfying the equation, particularly since it represents a conic passing through the origin. The transformation of the equation into a standard form is mentioned, emphasizing that the origin is an inherent solution. While the nature of the solution set can be influenced by the coefficients, there are limitations on extracting specific restrictions from the coefficients themselves. Overall, the conversation highlights the complexity of analyzing the equation while acknowledging the fundamental role of the origin in its solutions.
Khan86
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Given the general equation Ax^2 + Bxy + Cy^2 = 0, my question is what kind of restrictions can you put on A, B, and C such that the equality holds?
 
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The equation is that of a conic passing through the origin (you can see that x = 0, y = 0 is a solution). It is a quadratic in x and y and if you set a precondition that x and y are nonzero, you can get it as a quadratic in x/y or y/x. Don't do that though :-p
 
Khan86 said:
Given the general equation Ax^2 + Bxy + Cy^2 = 0, my question is what kind of restrictions can you put on A, B, and C such that the equality holds?

I'm not sure I understand the question. Whatever values of A, B and C you choose there will be a set of points in the xy plane that make the statement true. Restricting the coefficients to specific ranges will affect the nature of the set of points that make the statement true.
 
Tide is correct. The most you can do is to think of it as a conic passing through the origin. The general equation of the second degree is

A'x^2 + 2H'xy + B'y^2 + 2G'x + 2F'y + C = 0

is easily transformed into yours using

A' = A
H' = B/2
G' = 0
F' = 0
C = 0

Since it passes through the origin **which is a glaring solution by the way for the simple reason that it is most obvious** we cannot possibly restrict the solution set to exclude x = 0, y = 0. Given that, if you now treat the equation SOLELY as a quadratic in x (or in y) you can get x as a function of y (or y as a function of x). You may be able to say something about the nonegative nature of the discriminant of the equation, but there is nothing much that can be extracted about the coefficients from it.

Cheers
Vivek
 
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