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

  • Thread starter Sven
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In summary, the set of all (x,y) satisfying the given equation can take on different forms depending on the values of A and C. If A and C are both zero, the equation simplifies to a line. If only one of them is zero, it simplifies to a parabola. If both are nonzero, it can be a circle, ellipse, or hyperbola, depending on the values of A and C. The presence or absence of an xy term also plays a role in determining the geometric shape of the equation.
  • #1
Sven
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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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  • #2
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
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
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
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.
 
  • #6
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.
 

What is the general form of the equation Ax^2 + Bx + Cy^2 + Dy + E = 0?

The general form of this equation is a quadratic equation, with the variables x and y raised to the second power. It also includes coefficients A, B, C, D, and a constant term E.

What do the terms A, B, C, D, and E represent in this equation?

A, B, C, D, and E are all coefficients in the quadratic equation. A and C are the coefficients of the squared variables x^2 and y^2, respectively. B and D are the coefficients of the linear variables x and y, respectively. E is a constant term.

How do I solve for x and y in this equation?

This equation can be solved using the quadratic formula or by factoring. Once you have solved for x, you can substitute the value of x back into the original equation to solve for y.

What is the discriminant of this equation and how does it affect the solutions?

The discriminant of this equation is b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation. This value determines the nature of the solutions of the equation. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. If it is negative, there are no real solutions.

Can this equation be graphed?

Yes, this equation can be graphed on a two-dimensional coordinate plane. The x and y variables represent the x-axis and y-axis, respectively. The solutions of the equation will be represented by the points where the graph intersects the x-axis.

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