# Ax^2+Bxy+Cy^2 = 0 question

1. Oct 14, 2004

### Khan86

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?

2. Oct 14, 2004

### maverick280857

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 :tongue2:

3. Oct 14, 2004

### Tide

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.

4. Oct 15, 2004

### maverick280857

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