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Homework Help: Conditions on the coefficients of conics equations

  1. Apr 3, 2005 #1
    I have been taught that the equation [tex] Ax^2 + Cy^2 +Dx + Ey +F = 0 [/tex] represents a general form of conics.

    Then the conditions of the coefficients in the equation could identify which type of conics the equation represents...

    Circle: A=C
    Ellipse: A does not=C and AC>0
    Hyperbola: AC<0 and if the coefficients have opposite signs
    Parabola: A=0 OR C=0

    The thing that I do not understand is why... I was wondering if anyone knows a way to explain the reasons to me?
  2. jcsd
  3. Apr 3, 2005 #2


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    I didn't knew about that but it's very interesting. Let's prove it for the circle. We are used to the equation of a circle of radius R centered at (a, b) to be

    [tex](x-a)^2+(y-b)^2 = R^2[/tex]


    Ok, well if A = C, our cameleonic forumla becomes

    [tex]Ax^2 + Ay^2 + Dx + Ey + F = 0 [/tex]

    [tex]x^2 + y^2 + \frac{D}{A}x + \frac{E}{A}y + \frac{F}{A} = 0[/tex]

    Now notice that

    [tex](x + D/2A)^2 = x^2 + \frac{D}{A}x + (D/2A)^2[/tex]


    [tex](y + E/2A)^2 = y^2 + \frac{E}{A}y + (E/2A)^2[/tex]


    [tex]x^2 + y^2 + \frac{D}{A}x + \frac{E}{A}y + \frac{F}{A} = (x + D/2A)^2 + (y + E/2A)^2 - (D/2A)^2 - (E/2A)^2 + F/A = 0[/tex]

    [tex](x + D/2A)^2 + (y + E/2A)^2 = (D/2A)^2 + (E/2A)^2 - F/A[/tex]

    This is the equation of a circle centered at (-D/2A, -E/2A) and whose radius is [itex]\sqrt{(D/2A)^2 + (E/2A)^2 - F/A}[/itex]

    The other proofs are probably similar.
  4. Apr 3, 2005 #3
    :bugeye: whoa... You're really smart to be able to get all of those... but I'm very sorry to say that I don't think I understand it. When the tutor explained it was very simple, it's just that when I got home and started on the hw that I got all confused... :cry: And since I'm only starting on the conics stuff I don't think I would be able to go into more equation proofs. If it's not too rude to ask would it be alright if I get a more simple reason? Say for example the one that I know would be the parabola:

    Since x-h=a(y-k)^2 is the equation for the vertical parabola, A in that conics equation would have to equal to 0 because there is no x^2 when you expand x-h=a(y-k)^2.

    And since y-k=a(x-h)^2 is the equation for a horizontal parabola, C in that conics equation would have to equal to 0 because there is no y^2 when you expand y-k=a(x-h)^2.

    I hope it's not too much of a problem I hope to get something simple like that.. and sorry again ><!!
  5. Apr 3, 2005 #4
    For a parabola, only one squared term is present, either on the x or y. The fact that there is only one squared term in that general form is what generates the parabola, which could be considered a function if the parabola is not turned on its side.

    Think of the other three conic sections as consisting of two parts which open either towards each other (circle, ellipse) due to the x^2 and y^2 terms having the same signs or away from each other (hyperbola) due to the x^2 and y^2 terms having opposite signs. Since there are two parts to the other three conic sections, the equation that represent either of those three conic sections can't be considered as functions.
  6. Apr 4, 2005 #5


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    Strictly speaking [tex] Ax^2 + Cy^2 +Dx + Ey +F = 0 [/tex] is the general form for a conic section with axes parallel to the coordinate axes. A general conic can have axes at angles to the coordinate axes- that will give the general formula
    [tex] Ax^2 + Bxy+ Cy^2 +Dx + Ey +F = 0 [/tex]
    (Didn't you wonder why they skipped over B?)
  7. Apr 4, 2005 #6
    ooh~ those helped me understand more ^^. Thank you!
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