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How reconstitute a conic equation by 4 roots?

  1. Sep 29, 2015 #1
    Given two points in x-axis, a and b, is possible formulate a quadratic equation whose roots intersects a and b: (x-a)(x-b)=0

    So, is possible make the same with a conic equation? Given 4 points, two in x-axis (a and b) and two in y-axis (c and d), is possible reconstitute a conic equation whose roots in x and y are a, b, c and d?
  2. jcsd
  3. Sep 29, 2015 #2


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    Not uniquely. This is [itex]x^2- (a+ b)x+ ab= 0[/itex]. Multiplying the entire equation by any non-zero number gives another quadratic equation having the same roots.
    If you want the quadratic function with leading coefficient 1, given by [itex]y= x^2+ bx+ c[/itex] then you need to determine two values, a and b. That requires two linear equations which you can get by replacing x and y with values of two points, not just where the graph crosses the x-axis.

    A general conic is of the form [itex]ax^2+ bxy+ cy^2+ dx+ ey+ f= 0[/itex]. Even assuming a= 1 we have five coefficients, b, c, d, e, and f, to determine. That will require five points, not just four.
  4. Sep 29, 2015 #3
    I want ANY conic equantion that intersects a, b, c and d! And, more importante: I want to write this conic in the format (x-a)(x-b) + (y-c)(y-d) + (x-a)(x-d) = 0! Or, in some format like this...
  5. Sep 29, 2015 #4


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    Since you know what "conics" are, surely you know that the graphs of conics are one of:
    a single straight line,
    two parallel straight lines,
    two intersecting straight lines.
    (The last three are "degenerate conics")

    NONE of those cross the x-axis (or any horizontal straight line) more than twice so what you are asking for is impossible.
  6. Sep 29, 2015 #5


    Staff: Mentor

    I don't believe HallsOfIvy read your question correctly. Given two x-intercepts a and b, and two y-intercepts c and d, there are many conics that meet this requirement: circles, ellipses, and hyperbolas. The only conic that doesn't satisfy the requirements is a parabola.
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