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Parabolas and hyperbolas - how do you find them?

  1. Jun 6, 2008 #1
    Finding linear equations for initial conditions has never been a problem from me since is easy to see (x1, y1) and (x2, y2) can be plugged into the slope equation. But what if I'm looking for a parabolic or even a hyperbolic line, how do I then find its equation with the initial condition that it must have some (x,y) and (x2, y2) points?
     
  2. jcsd
  3. Jun 6, 2008 #2

    mathman

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    The general form for conic sections (parabolas, hyperbolas, ellipses, circles, two lines) looks like ax2+bxy+cy2+mx+ny+k=0. Since you can divide by any constant, there are five parameters. Therefore you need to specify the curve at five points.
     
  4. Jun 6, 2008 #3
    Well, given that I still don't see a way to make sense out of that general equation since conic secs don't have an actual slope (unless one were to find its many tangents). I'm convinced that there is a way to solve for a parabolic equation given two sets of coordinates though
     
  5. Jun 6, 2008 #4
    You plug the coordinates into the equation for each of the 5 points. That gives you 5 equations and 5 unknowns, which you can solve for.

    Eg: suppose the curve passes through the following points (im just making these points up):
    (1,2); (0,0); (3,4); (-1,3); (-2,-3).

    Plugging each point into the equation
    [tex]x^2+b \cdot x \cdot y+c \cdot y^2+m \cdot x+n \cdot y+k=0[/tex],

    you get the following 5 equations:
    [tex]1^2+b \cdot 1 \cdot 2+c \cdot 2^2+m \cdot 1+n \cdot 2+k=0[/tex]
    [tex]0^2+b \cdot 0 \cdot 0+c \cdot 0^2+m \cdot 0+n \cdot 0+k=0[/tex]
    [tex]3^2+b \cdot 3 \cdot 4+c \cdot 4^2+m \cdot 3+n \cdot 4+k=0[/tex]
    [tex](-1)^2+b \cdot (-1) \cdot 3+c \cdot 3^2+m \cdot (-1)+n \cdot 3+k=0[/tex]
    [tex](-2)^2+b \cdot (-2) \cdot (-3)+c \cdot (-3)^2+m \cdot (-2)+n \cdot (-3)+k=0[/tex]

    Simplifying,
    [tex]1+2 b+4 c+m +2 n+k=0[/tex]
    [tex]k=0[/tex]
    [tex]9+12 b+16 c+3 m+4 n+k=0[/tex]
    [tex]1-3 b +9 c-1 m+3 n+k=0[/tex]
    [tex]4+6 b+9 c-2 m-3 n+k=0[/tex]

    Now you can solve these equations for b,c,m,n, and k.

    ----

    I went ahead and solved this system of equations since it isn't much trouble, and got the following answers:
    b = 116/33
    c = -95/33
    m = -133/11
    n = 257/33
    k = 0

    So,
    [tex]x^2+\frac{116}{33}x \cdot y-\frac{95}{33} y^2-\frac{133}{11} x+\frac{257}{33} y=0[/tex]

    Plotting the resulting figure, it looks like a hyperbola, and it passes through all the points we want. Whew!
    http://img520.imageshack.us/img520/8662/hyperbolapl5.png [Broken]
     
    Last edited by a moderator: May 3, 2017
  6. Jun 7, 2008 #5
    Sorry that you had to go through all that trouble maze. Thanks for the example
     
  7. Jun 8, 2008 #6

    HallsofIvy

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    Then you are convinced wrong. y= x2 and y= 2x2- 1
    both pass through the two points (1, 1) and (-1, 1).

    Even given the information that the graph is a parabola, concave upward, with vertical axis, the two points (1, 1) and (-1, 1) are not enough to define a specific parabola.
     
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