# Ax^2 + Bx + Cy^2 + Dy + E = 0 help?

1. Oct 23, 2009

### Sven

1. The problem statement, all variables and given/known data

Suppose that A and C are not both 0. Show that the set of all (x,y) satisfying

Ax^2 + Bx + Cy^2 + Dy + E = 0

is either a parabola, an ellipse, or an hyperbola (or a degenerate case: two lines, one line, a point, or 0.)

Now consider separately the cases where A and B are both positive or negative, and where one is positive while theo ther is negative. When do we have a circle?

2. Relevant equations

3. The attempt at a solution

Okay I don't know how to show this. Do I have to cut it up into 20-30 cases and prove for stuff like when A is not 0 and B is not 0 and everything else is 0, and when C is not 0 and D is not 0 and everything else is 0?

2. Oct 23, 2009

### Staff: Mentor

You have way too many cases. You only need to worry about the values for A and C.
Here are the cases you need to explore. (BTW, "!=" means "not equal".)
1. A = 0 and C != 0
2. A != 0 and C = 0
3. A and C both nonzero, broken up into the following subcases:
...a. A > 0 and C > 0
...b. A < 0 and C < 0
...c. A > 0 and C < 0
...d. A < 0 and C > 0

3. Oct 23, 2009

### lanedance

though as you could essentially interchange x & y in any solution, the groupings (1 & 2), (3a & 3b), (3c&3d) essentially give the same form of solution further reducing the number of cases you have to deal with

4. Oct 23, 2009

### Sven

oh okay, I see, thanks.

so for the cases I do have... can someone tell me if I'm doing these right? I'm getting stuck on some.

If A and B are nonzero, then Ax2 + Bx = 0. Divide by x, then divide by a, and you get x = -B/A. This is a line. Is that right?

So if A and C are nonzero, Ax2 + Cy2 = 0. How would I do this one?

5. Oct 23, 2009

### Staff: Mentor

No, you're really on the wrong track. In your first one, there is no case for B being zero or nonzero.

Go down the list of cases that I gave you and determine the result of that assumption. For example, in the first case I gave, A = 0 and C != 0. Nothing is assumed about any other coefficient. What does your equation simplify to in this case? And what sort of geometric object is described in this case.

Do the same for each case.

6. Oct 24, 2009

### lurflurf

There is no xy term? That is where all the fun is! The standard way is either to complete the square or use the discriminant (after having proven it first) which in your notation is AC.