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Normal line question

  1. Apr 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the normal line to the curve y(sqrd) - x(sqrd)y + 3 = 0 at the point (-2,3). (standard form)



    2. Relevant equations
    y - y = m(x - x)


    3. The attempt at a solution

    dy
    __ = -6
    dx

    y - 3 = -6(x + 2)
    6x + y +9 = 0
     
  2. jcsd
  3. Apr 12, 2007 #2
    What you have found is the equation of the tangent line at (-2,3). You need to use the same method, but use the slope of the normal to the curve at that point. Do you know how the slopes of two mutually perpendicular lines are related to each other?
     
  4. Apr 12, 2007 #3
    Is it the reciprocal?

    1
    _ ?

    6
     
  5. Apr 12, 2007 #4

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    Close, but not quite. You should look it up.
     
  6. Apr 12, 2007 #5
    its not the reciprocal?

    my current answer in standard form is..

    x + 6y - 16 = 0
     
    Last edited: Apr 12, 2007
  7. Apr 12, 2007 #6

    daniel_i_l

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    Gold Member

    What about the sign? (though it seems as though you've taken that into consideration with 1/6)
     
  8. Apr 12, 2007 #7
    I did, should be -1/6 typo above. sorry

    Is it correct now?
     
  9. Apr 12, 2007 #8
    The slope of the normal is 1/6. The equation of the normal is x-6y+20=0
     
  10. Apr 12, 2007 #9
    thanks for the help. Just wanted to verify one thing though, dont you muliply -1 by the (x + 2) making them both negative.

    y - 3 = -1/6 (x + 2)
    6y - 18 = -x -2
    x + 6y -16 = 0
     
  11. Apr 12, 2007 #10
    No, you don't.

    (y-y0) = m(x-x0)

    y0 = 3, x0 = -2, m = 1/6.
     
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