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Math Help - Maximum number of intersections of two quadratics

  1. #1
    r45
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    Maximum number of intersections of two quadratics

    Hi there,

    I am wondering what is the maximum number of points of intersection of 2 quadratics of the form:

    ax^2 + bxy + cy^2 + dx + ey + f = 0

    I think the answer is 4, because expressing y in terms of x has a \pm sign, and then subbing each of these y-values in to the other quadratic leads to 2 solutions each through use of the quadratic formula. Thus 2 x 2 = 4 is the maximum number of intersections.

    This also seems intuitive, if we imagine for example a hyperbola being intersected 4 times by an ellipse. I can't picture in my head any 2 quadratics who intersect more than 4 times.

    Anyway if someone could please verify or disprove my assertion I'd be very grateful!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by r45 View Post
    Hi there,

    I am wondering what is the maximum number of points of intersection of 2 quadratics of the form:

    ax^2 + bxy + cy^2 + dx + ey + f = 0

    I think the answer is 4, because expressing y in terms of x has a \pm sign, and then subbing each of these y-values in to the other quadratic leads to 2 solutions each through use of the quadratic formula. Thus 2 x 2 = 4 is the maximum number of intersections.

    This also seems intuitive, if we imagine for example a hyperbola being intersected 4 times by an ellipse. I can't picture in my head any 2 quadratics who intersect more than 4 times.

    Anyway if someone could please verify or disprove my assertion I'd be very grateful!
    infinite number of points take the same quadratic for the two curves
    if the two curves are different you have 4 points

    ax^2 + bxy + cy^2 + dx + ey + f = 0

    if we take two curves say

    a_1x^2 +b_1xy + c_1 y^2 + d_1 x + e_1 y + f_1 = 0 ...(1)

    a_2x^2 +b_2xy + c_2 y^2 + d_2 x + e_2 y + f_2 = 0 ...(2)

    rewrite the first one like

     c_1 y^2+ e_1 y +b_1xy + d_1 x +a_1x^2 + f_1 = 0

    c_1 y^2 + y(e_1 + b_1x) + d_1x + a_1x^2 +f_1 = 0

    solve for y

     y = \frac{-(e_1 + b_1x) \pm \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}

    now we have to curves from (1)

    y_1 = \frac{-(e_1 + b_1x) + \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}

    y_2 = \frac{-(e_1 + b_1x) - \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}

    and these two curve dose not intersect with each other you can see that

    do the same thing for (2) you will have two curves

    y_3 =\frac{-(e_2 + b_2x) + \sqrt{(e_1 + b_1x)^2 - 4c_2( d_2x + a_2x^2 +f_2)}}{2c_2}

    y_4 = \frac{-(e_2 + b_2x) - \sqrt{(e_2 + b_2x)^2 - 4c_2( d_2x + a_2x^2 +f_2)}}{2c_2}

    and these two never intersect
    Last edited by Amer; April 6th 2010 at 11:21 AM.
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