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Finding the Point of Tangency Theoretically

  1. Mar 30, 2006 #1
    Hey I was given the question:
    A (x1,y1) and B (x2,y2) are two points on the parabola y=ax^2+bx+c. At what point is the tangent to the parabola parallel to the secant AB.
    Here are the steps I took, im just wanting to know if im heading in the right direction.

    First I just made the slope of AB = m.
    then found the derivative of ax^2+bx+c (2ax+1) and made m=2ax+1.
    Then isolated x, which would give an x coordinate of m/2a+1.
    Took the x value and subed it into the equation ax^2+bx+c to get a y value.

    Is this how you would go about this question?? thanx
     
  2. jcsd
  3. Mar 30, 2006 #2
    Why 2ax + 1, instead of 2ax + b ?
     
  4. Mar 30, 2006 #3
    :smile: ahh, yes good point, i wasn't thinking. Thanx for catching that PPonte!
     
  5. Apr 1, 2006 #4

    arildno

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    Nowhere have you related your "x" to the x1 and x2; what you have done, is to parametrize the parabola in terms of the local slope m which, by the way, you haven't done correctly.

    To continue along your track:
    In general, you'll get for the tangent slope m
    [tex]m=2ax+b\to{x}=\frac{m-b}{2a}[/tex]

    If you now can find the m that is the slope of the secant between x1 and x2, you are done.
     
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