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Tide

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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.Khan86 said:

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[tex]A'x^2 + 2H'xy + B'y^2 + 2G'x + 2F'y + C = 0[/tex]

is easily transformed into yours using

[tex]A' = A[/tex]

[tex]H' = B/2[/tex]

[tex]G' = 0[/tex]

[tex]F' = 0[/tex]

[tex]C = 0[/tex]

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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