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Need quick help

  1. May 4, 2008 #1
    i have 2 simmilar questions
    1) Find the equation of the line tangent to the curve at the point
    xy + y = 2 ; (1, 1) and i did this

    xy + y = 2
    y + xy' + y' = 0
    xy' + y' = - y
    y'(x + 1) = -y
    y' = -y/(x + 1)
    y' = -1/2
    y- y_1 = m(x - x_1) and i hope so this is OK
    but how can i do this one???
    2) xy + x = 2 ; (1, 1) please help have to answer in 3 hours
  2. jcsd
  3. May 4, 2008 #2


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    Homework Helper

    Well you found y' at (1,1) [your "m"]. In that form [itex]y- y_1 = m(x - x_1)[/itex], [itex](x_1,y_1)[/itex] is a point on the line...which you have.

  4. May 5, 2008 #3


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    Staff Emeritus
    Science Advisor

    That's good so far but what you give in your final line is a general formula, not the answer to this problem. What is your answer?

    ?? You have these labled (1) and (2) so I thought they were different problems- but they are exactly the same. Are you just asking HOW to give the final answer? Use what you said above: y-y_1= m(x- x_1). (x_1, y_1) is, of course, the point you were given: (1, 1) and m is the slope you calculated: -1/2. Put those numbers in.
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