1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General Quadratic Equation

  1. Jul 9, 2009 #1
    The general quadratic equation:

    Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

    How is this derived? How does the discriminant, B2 – 4AC determine the class of the conic section?
  2. jcsd
  3. Jul 9, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you asking how B2-4AC is derived? Because the actual polynomial equation isn't derived, or something that you prove, it just is.
  4. Jul 9, 2009 #3


    User Avatar
    Science Advisor

    That equation, [itex]Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ F= 0[/itex] is the "general quadratic equation" because you cannot have any other terms in a polynomial without introducing higher powers- those are all the terms of degree two or less. As for determining the "discriminant", there are several ways to do that, essentially "rotating" the coordinate system to get rid of the "xy" term. If you let [itex]x= x' cos(\thet)- y' sin(\theta)[/itex] and [itex]y= x'sin(\theta)+ y'cos(\theta)[/itex] and substitute those into the equation, you get a very complicated equation in x' and y' with coefficients that depend on [itex]\theta[/itex]. Choose theta so that the coefficient of x'y' is 0 and see what kind of conic you have. You really only need to look at the "quadratic" part- you can ignore D, E, and F.

    A simpler calculation, but more "sophisticated", is to write the quadratic part of the equation as a "bilinear tranformation":
    [tex]\begin{bmatrix} x & y\end{bmatrix}\begin{bmatrix} A & \frac{B}{2} \\ \frac{B}{2} & C \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}[/tex]
    and look at the eigenvalues and eigenvectors of that 2 by 2 matrix. It will be a circle the two eigenvalues are the same, an ellipse if they are different but of the same sign, a parabola if one of them is 0, and an hyperbola if they are of different signs.
  5. Jul 9, 2009 #4
    What do you mean exactly by "rotating" the graph? Are you referring to an xy or xyz plane graph?
  6. Jul 10, 2009 #5


    User Avatar
    Science Advisor

    Your equation only had x and y so, of course, I was referring to an xy graph. And by "rotating" I mean moving the axes so that the origin remains fixed while the two axes move through an angle [itex]\theta[/itex]. For example, rotationg by [itex]\pi/4[/itex] radians, the x-axis moves to the line y= x and the y-axis to the line y= -x.
  7. Jul 10, 2009 #6
    The product [itex]xy[/itex] from [itex]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/itex] will make the conic oblique to the coordinate axes. The substitution [itex]x = x' \cos(\theta) - y' \sin(\theta)[/itex] and [itex]y = x' \sin(\theta)+ y' \cos(\theta)[/itex] allow us to make the conic parallel or perpendicular to one of the coordinate axes. Talking about the discriminant B2 - 4AC, if [itex]B^2 < 4AC[/itex] the conic is an ellipse, if [itex]B^2 = 4AC[/itex] conic is parabola, and if [itex]B^2 > 4AC[/itex] the conic is hyperbola. On how it exactly determine the class of conic section, I don't have an answer for now. But one thing I am certain that the discriminant defines the conic similar to how the eccentricity defines it., ie e < 1 (ellipse), e = 1 (parabola), and e < 1 (hyperbola).
    Last edited: Jul 10, 2009
  8. Jul 10, 2009 #7


    User Avatar
    Science Advisor

    The best way to see that is the "eigenvalue" method I mentioned.
    The determinant of the matrix
    [tex]\begin{bmatrix}A & \frac{B}{2} \\ \frac{B}{2} & C\end{bmatrix}[/tex]
    is AC- B2/4= (1/4)(4AC- B) and will be positive if B- 4AC< 0, negative if B- 4AC> 0, and 0 if B- 4AC= 0. The product of the eigenvalues is equal to the derivative so we have that the two eigenvalues are of the same sign (a circle or ellipse) if B2- 4AC< 0, of opposite sign (a hyperbola) if B2- 4AC> 0, and at least one 0 (a parabola) if B2- 4AC= 0.

    (Two be complete, you should also include "degenerate conics": if A= 1, C= -1, B= D= E= F= 0, the equation is x2- y2= (x- y)(x+ y)= 0 which graph is the two lines x+ y= 0 and x- y= 0 (and is a "degenerate" hyperbola). If A= 1, B= -2, C= 1, D= E= F= 0, the equation is x2- 2xy+ y2= (x- y)2= 0, a single straight line (a "degenerate" parabola). If A= 1, C= 1, B= D= E= F= 0, the equation if x2+ y2= 0, the single point (0, 0) (a "degenerate" circle).)
  9. Jul 11, 2009 #8
    I wasn't sure. Is it true, for the quadratic equation, that ax2 + bx + c = 0 = f(x)? I was just thinking that Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 = f(x,y). I don't know; I may be missing something. The graph of a quadratic equation is by the function f(x) = ax2 + bx + c. I just don't understand what everything implies and why y-values are suddenly intermingled with x-values into the general quadratic equation.
    I would also like to understand why B2 = 4AC
  10. Jul 11, 2009 #9


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    You can achieve enough justification for yourself using the Distance Formula to derive the equations for all of the conic sections according to their Distance Formula definitions. You should at least intuitively understand that switching the roles of x and y will give you graph results tilted by a Right angle. You should now see that each of the terms in the Quadratic general equation are accounted. The XY term occurs because of rotation of the graph. Justifying this precisely is beyond my own understanding, but it would involve Trigonometry. For an improved understanding of Y and Y2 terms, when the X terms are also present (or even when they are not), you can complete the square for them in order to help obtain a standard form equation or function.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook