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Find tangent lines to both curves

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equation of all straight lines, if any, that are tangent to both the curves [itex]y = {x^2} + 4x + 1[/itex] and [itex]y = - {x^2} + 4x - 1[/itex].


    2. Relevant equations



    3. The attempt at a solution
    Suppose such a line exists and its slope is m. Let [itex]({x_1},{y_1})[/itex] and [itex]({x_2},{y_2})[/itex] be the tangent points on the curves [itex]y = {x^2} + 4x + 1[/itex] and [itex]y = - {x^2} + 4x - 1[/itex] respectively.
    Then [itex]{y_1} = {x_1}^2 + 4{x_1} + 1[/itex] and [itex]{y_2} = - {x_2}^2 + 4{x_2} - 1[/itex].
    The slope of the curves at [itex]{x_1}[/itex] is [itex]2{x_1} + 4[/itex] and at [itex]{x_2}[/itex] is [itex]-2{x_2} + 4[/itex]. Thus [itex]m=2{x_1} + 4=-2{x_2} + 4 \Rightarrow {x_1} = - {x_2}[/itex]. Moreover, [itex]m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{( - {x_2}^2 + 4{x_2} - 1) - ({x_1}^2 + 4{x_1} + 1)}}{{{x_2} - {x_1}}} = \frac{{{y_1}}}{{{x_1}}}[/itex].
    What should be the next steps here? Thanks!
     
  2. jcsd
  3. Oct 2, 2012 #2

    SammyS

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    You have it nearly done.

    Well, you have:
    [itex]x_1=-x_2[/itex]

    [itex]m=-2{x_2} + 4[/itex]

    [itex]\displaystyle m=\frac{{( - {x_2}^2 + 4{x_2} - 1) - ({x_1}^2 + 4{x_1} + 1)}}{{{x_2} - {x_1}}}[/itex]​

    Plug -x2 in for x1 in that last equation & equate that to your other expression for m that has only x2 in it.

    Solve for x2.
     
  4. Oct 2, 2012 #3
    Gosh I am so blind, thank you so much!
     
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