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Deriving Equations of a Parabola

  1. May 29, 2008 #1
    Hey,

    1. The problem statement, all variables and given/known data
    How do you derive the equations of the parabola from the general equation of a Conic Section?

    2. Relevant equations
    General Equation of a Conic Section,
    [tex]
    {{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}
    [/tex]

    Where (for a parabola),
    [tex]
    {{\{}A,B,C,D,E,F{\}}}{{\,}{\,}{\,}}{\in}{{\,}{\,}{\,}}{\mathbb{R}}
    [/tex]
    [tex]
    {{\{}A, C{\}}}{{\,}{\,}{\,}}{\neq}{{\,}{\,}{\,}}{0}
    [/tex]
    [tex]
    {{{B}^{2}}-{{4}{A}{C}}} = {0}
    [/tex]

    3. The attempt at a solution
    From the general equation of a Conic Section,
    [tex]
    {{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}
    [/tex]

    How do I derive the following formulas for a parabola:
    General Form for a Parabola,
    [tex]
    {{{{\left(}{{{H}{x}}+{{I}{y}}}{\right)}}^{2}}+{{J}{x}}+{{K}{y}}+{L}} = {0}
    [/tex]
    Where,
    [tex]
    {{{I}^{2}}-{{4}{H}{J}}} = {0}
    [/tex]

    Analytic Geometry Equations,
    Vertical Axis of Symmetry
    [tex]
    {{{\left(}x-h{\right)}}^{2}} = {{{4}{p}}{{{\left(}y-k}{\right)}}}
    [/tex]
    [tex]
    {y} = {{{a}{{x}^{2}}}+{{b}{x}}+{c}}
    [/tex]
    Horiztonal Axis of Symmetry
    [tex]
    {{{\left(}y-k{\right)}}^{2}} = {{{4}{p}}{{{\left(}x-h{\right)}}}}
    [/tex]
    [tex]
    {x} = {{{d}{{y}^{2}}}+{{e}{y}}+{f}}
    [/tex]

    I read that every parabola is a combination of transformations of the parabola, [tex]{{y}={{x}^{2}}}[/tex]; but I'm not quite sure how that helps.

    Thanks,

    -PFStudent
     
    Last edited: May 29, 2008
  2. jcsd
  3. May 29, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm not precisely sure of what exactly you are supposed to do, but the only way that the quadratic terms Ax^2+Bxy+Cy^2 can be factored as (Hx+Iy)^2 is that B^2-4AC=0. Just multiply (Hx+Iy)^2 out and identify the terms. Does that help?
     
  4. May 30, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    First you need to find the "principal axes" (axis of symmetry and perpenpendicular to it through the vertex) and rotate the coordinate system so that new x' and y' axes are the principal axes.

    There are two ways to do that, basically using the quadratic terms [itex]Ax^2+ Bxy+ Cx^2[/itex] (By the way, you don't have to put braces, { }, around every term. They are only necessary when you want an entire expression in a particular place. For example, [i t e x]e^{xy+ b}[/i t e x] gives [itex]e^{xy+ b}[/itex] while [i t e x]e^xy+ b[/i t e x] gives [itex]e^xy+ b[/itex].)

    Rotating the axes by angle [itex]\theta[/itex], so that the new x'y'- axes are at angle [itex]\theta[/itex] to the old xy- axex, x and y are given, in terms of the new x', y' variables, by [itex]x= x'cos(\theta)+ y'sin(\theta)[/itex] and [itex]y= -x'sin(\theta)+ y'cos(\theta). Replace x and y by those in [itex]Ax^2+ Bxy+ Cx^2[/itex] and choose [/itex]\theta[/itex] so that the coefficient of x'y' is 0. Use those formulas with the correct value of [itex]\theta[/itex] to replace x and y in the entire formula. If it really is a parabola, both the x'y' and y'2 terms should vanish.

    The other way is to write it as a matrix problem:
    [tex]Ax^2+ Bxy+ Cy^2= \left[\begin{array}{cc}x & y \end{array}\right]\left[\begin{array}{cc}A & \frac{B}{2} \\ \frac{B}{2} & C\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right][/tex]

    You can "diagonalize" that matrix (so writing the quadratic without the xy term) by finding the eigenvalues and eigenvectors of the coefficient matrix. The eigenvectors will point along the principal axes. Since this is a parabola, one of the eigenvalues should be 0.
     
  5. May 31, 2008 #4
    Hey,

    Thanks for the replies: Dick and HallsofIvy.

    Yea, to be honest I had not seen that general form for a parabola before until I saw it on wikipedia, maybe it is wrong? - here is the link, (Scroll down a bit to the section "General Parabola").

    http://en.wikipedia.org/wiki/Parabola

    Let me know what you think.

    Hmm, ok that makes sense, I will try that. However, I thought since a parabola - in its' most general form - could be expressed as a conic section with the constraint that [tex]{{B^2} = {4AC}}[/tex]. Then shouldn't I be able to derive the more specific forms of a parabola from the general equation of a conic section with the constraint that [tex]{{B^2} = {4AC}}[/tex]?

    Where the general equation of a Conic Section is,
    [tex]
    {{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}
    [/tex]

    Also, yea I tend to use a bit too many braces ({ }), just a habit from first learning [itex]LaTeX[/itex] way back. :redface:

    Thanks,

    -PFStudent
     
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