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

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Homework Help Overview

The problem involves analyzing the equation Ax² + Bx + Cy² + Dy + E = 0, where A and C are not both zero. Participants are tasked with determining the geometric nature of the solutions (parabola, ellipse, hyperbola, or degenerate cases) based on the values of the coefficients A, B, C, and D.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants question the necessity of considering multiple cases based on the values of A and C, suggesting a reduction in the number of cases by recognizing interdependencies. Others express uncertainty about the implications of specific coefficient values on the resulting geometric shapes.

Discussion Status

Participants are actively engaging with the problem, exploring different cases and questioning their assumptions. Some guidance has been offered regarding the simplification of cases, but there is no explicit consensus on the approach to take for each scenario.

Contextual Notes

There is a focus on the conditions under which the equation describes different conic sections, with participants noting the absence of an xy term and discussing methods such as completing the square or using the discriminant.

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Homework Statement



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?

Homework Equations





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?
 
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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
 
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
 
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?
 
Sven said:
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?
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.
 
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.
 

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