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Finding tangent lines that pass through given points

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the co-ordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0)


    2. Relevant equations
    f '(x) = nxn-1


    3. The attempt at a solution
    I am really not sure how to attack this question. My initial thoughts are to find f '(x) then, given the points (a,0), create a point-slope equation for a line. But, this would only give me one equation. On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point. I'm not sure how I would find the co-ordinates either.

    f '(x) = 3x2

    y = 3x2 (x-a)
     
  2. jcsd
  3. Feb 9, 2012 #2

    SammyS

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    ... and y = x3 .

    Hello methionine. Welcome to PF !
     
  4. Feb 9, 2012 #3
    I'm sorry, what do you mean? ".....and y = x3
     
  5. Feb 10, 2012 #4

    SammyS

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    Your line passes through the point (x, x3) as well as the point (a, 0).
     
  6. Feb 10, 2012 #5

    hotvette

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    I believe there are indeed two lines tangent to the curve that go through the point (a,0). Look carefully.
     
  7. Feb 13, 2012 #6
    Hey guys, I spent the last day mulling this question over in my head, and decided to attack it in a slightly different way.

    Basically, did what I was doing up to this point, set a point-slope form equation using x and x and x3 and x1 and y1....

    y-x3 = 3x2(x-x3)

    Now, I know the line has to pass through (a,0) as well, so I plugged those values into X and Y and ended up with two points.

    -x3 = 3x2(a-x3)

    Kind of an interesting question for me. I hope my work checks out!
     
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