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Finding the slope of line tangent to a parabola

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equations of both lines through the point (2,-3) that are tangent to the parabola y=(x^2)+x

    2. Relevant equations



    3. The attempt at a solution

    Took the derivative and got a slope of 5 and the slope of the normal line being -1/5, but the answer was marked wrong. How do I do this?

    Two equations I got
    y=-(1/5)x-(13/5)
    y=5x-13
     
  2. jcsd
  3. Sep 30, 2011 #2

    ElijahRockers

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    Choose a point (a,y(a)). A line that goes through this point, AND the given point must have a slope of [itex]\frac{y(a) - (-3)}{a - 2}[/itex] Also, the slope at point 'a' can be given by the derivative of the function. This gives you two equal expressions for the slope in terms of a. It will be a quadratic equation. The roots will be the x values at which the lines intersect the parabola.
     
    Last edited: Sep 30, 2011
  4. Sep 30, 2011 #3

    SammyS

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    What do you mean? "both lines"

    One of your lines is tangent to the parabola at (2, -3) .

    The other is normal to the parabola at (2, -3) .
     
  5. Sep 30, 2011 #4

    ElijahRockers

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    Actually I think it's tangent to the parabola at (2, 6)
     
  6. Sep 30, 2011 #5

    SammyS

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    Ha!

    Yup, the parabola doesn't pass through (2, -3) ! DUH
     
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