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Find the equation of the line tangent to two parabolas

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the two parabolas: [itex] f(x) = x^2 - 2x + 2[/itex] and [itex] g(x) = -x^2 - 2x - 2[/itex]. Find the equation of a line that is tangent to both curves.


    2. Relevant equations

    The given parabolas, equation for a line [itex]y = mx + b[/itex], and the derivatives of the two parabolas [itex]2x - 2[/itex] and [itex]-2x - 2[/itex]

    3. The attempt at a solution

    The line tangent to the two parabolas will pass through the points

    [itex](x_1 , y_1)[/itex]

    for the parabola [itex]f(x)[/itex] and

    [itex](x_2 , y_2) [/itex]

    for the parabola [itex]g(x)[/itex]

    so the equations for the two lines will be,

    for f' [itex]y_1 = (2x_1 - 2)x_1 +b [/itex]

    and

    g' [itex] y_2 = (-2x_2 - 2)x_2 + b[/itex].

    Since the slopes of both lines will be the same I thought that setting the two slopes equal might get me started so I wrote

    [itex]2x_1 - 2 = -2x_2 - 2 [/itex]

    which gave me [itex]\frac{x_1}{x_2} = -1 [/itex].

    I rearranged the two linear equations and set them equal:

    [itex]y_1 - (2x_1 - 2)x_1 = y_2 - (-2x_2 - 2)x_2[/itex]

    but that got me absolutely nowhere. I got the whole thing down to

    [itex]x^2 _1 + x^2 _2 = \frac{y_1 - y_2}{2}[/itex]

    but that doesn't help.

    I know I need to limit my variables and try to get the whole thing in terms of one variable but I'm at a loss as to how to make that happen. Every time I substitute and simplify I get either one or negative one which tells me nothing. I can't figure out how relate the equations and simplify. Any help is appreciated.
     
  2. jcsd
  3. Feb 13, 2013 #2
  4. Feb 13, 2013 #3

    haruspex

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    You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.
     
  5. Feb 13, 2013 #4
    I'm not sure what you mean. I accounted for them being on different parabolas with the subscripts but is there something in their being on different parabolas that I missed?
     
  6. Feb 13, 2013 #5
    And I just figured it out. That tutorial from UW did the trick which is kind of funny seeing that's where I did my undergrad. Thanks iRaid!
     
  7. Feb 13, 2013 #6

    haruspex

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    I mean that e.g. x1, y1 satisfy the equation of the first parabola.
     
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