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Geometry problem

  1. Jan 15, 2006 #1
    For the curve y = (1/3)x^3, given that 24x + 3y +2 = 0 is the equation of the tangent to the curve at the point (p,q) find p and q.

    I rearranged to get y = -8x - 2/3

    So the tangent gradient is -8

    I differentiated to get x^2 - 9

    Therefore x^2 - 9 = -8 and x = 1 or x = -1.

    From here how do I distinguish whether x = 1 or x = -1?

    Many thanks.
  2. jcsd
  3. Jan 15, 2006 #2
    for x =1
    substitute this into the equation of the tangent and the equation of the curve

    do the same thing for x = -1
    if in any of the two cases you get the y value that matches between the curve and the tangent then that point (x,y) lies on the curve, thus the line is tangent to that point on the curve
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