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Tangents and Parabolas

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    The line y = 4x-7 is a tangent to a parabola that has a y-intercept of -3 and the line x=1/2 as its axis of symmetry. Find the equation of the parabola.

    This is supposed to be done with regards to the discriminant.

    2. Relevant equations


    3. The attempt at a solution

    y=4x-7
    y-intercept = -7
    x-intercept = 7/4
    after drawing the graph, the parabola has to be concave up.

    integrating... (i have no idea what else to do)
    y=2x^2-7x+c
    discriminant = 49-4*2c
    = 49-8c
    =0
    8c=49
    c=49/8

    ...:confused:
     
  2. jcsd
  3. Sep 22, 2011 #2

    eumyang

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    Homework Helper

    No, you don't want to do that.

    The quadratic you're looking for is in the form of y = ax2 + bx + c. If (0, -3) is on the parabola, then you should know what c is.

    Since (0, -3) is on the parabola and the axis is x = 1/2, "reflect" the point over the x-axis and you'll find another point on the parabola. Check if this point is also on the given tangent line. (It is.)

    Plug this 2nd point into y = ax2 + bx + c. for x and y, and plug in the value you know for c, and now you have an equation in terms of a and b.

    Find the derivative of y = ax2 + bx + c, and use it, with the 2nd point and the knowledge of the slope of the given tangent line, to find a 2nd equation in terms of a and b. Now you'll have 2 equations and 2 unknowns. Solve for a and b.
     
  4. Sep 22, 2011 #3

    PeterO

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    Homework Helper

    The axis of symmetry x = 1/2 means the equation of the parabola is of the form

    y = a(x - 1/2)2 + c

    expand

    y = ax2 - ax + a/4 + c

    The y-intercept of -3 means (0,-3) is a point on the graph.
    Substituting that point into the formula says c + a/4 = -3
    That enables you to get c in terms of a, or a in terms of c, so your equation will have only one unknown in it.

    Knowing that y = 4x - 7 is a tangent, you know that when solving simultaneously, you will get only one solution [that is where the discriminant comes in].

    Do all that and you should find the equation of the parabola.
     
  5. Sep 22, 2011 #4

    NascentOxygen

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    Staff: Mentor

    This being precalculus, there may be a way to do it without differentiation.

    Let the parabola be y=ax2+bx+c
    and you can write out the general formula that we use to solve quadratics.
    The intercept and axis of symmetry allow you to determine some things about and between coefficients a, b, and c.

    Next, solve simultaneously to find the points of intersection of the parabola with the line y=4x-7. There are 2 general points of intersection, but you are told for this line & parabola these coincide as one. This information tells you that there is another discriminant you can equate to zero (or otherwise) to give more relationships among the 3 coefficients. I think there will now be ample information to determine all three.

    No calculus needed. :cool:

    Not that I have anything against the calculus. :smile: :!!) I love calculus!
     
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