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Tangent Line Problem

  1. Nov 5, 2006 #1
    Hi, I was wondering if anyone could help.

    Consider the curve given by: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

    I solved the derivative which came out to be (4x-2xy)/(x^2 + y^2 + 1

    1. Write an equation of each horizontal tangent line to the curve.
    2. The line through the origin with slope -1 is tangent ot the curve at point P. Find the x- and y- coordinates of point P.

    for 1 ) do I find when the derivative is 0? so I set the top = 0 to get y = 2.

    for 2) im not even sure how to start this.. so I have m = -1 and point (0,0) .. now what:confused:
  2. jcsd
  3. Nov 6, 2006 #2
    y = -x then is the line that is tangent to the original curve 2y^3 + (6x^2)y - 12x^2 + 6y = 1. -1 also equals that tangent, the derivative you obtained. The two equations - the derivative and the original curve - should have a common point, which is the point P you are looking for.
  4. Nov 6, 2006 #3


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    The line with m= -1, through (0,0) is, of course, y= -x. You now have two equations for the (x,y) coords of the point where that line is tangent to the curve: y'= (4x-2xy)/(x^2 + y^2 + 1)= -1 and the equation of the curve, 2y^3 + (6x^2)y - 12x^2 + 6y = 1. Knowing that y= -x should make that very easy. (If such a point exists.)
    Last edited: Dec 4, 2008
  5. Dec 2, 2008 #4
    can someone do out part 2 of this problem?
  6. Dec 2, 2008 #5


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    PF policy explicitly forbids doing homework for others. Just post any problems you encounter and we'll help.
  7. Dec 2, 2008 #6
    i dont understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?
  8. Dec 3, 2008 #7
    have you learnt the topic differentiation?
  9. Dec 3, 2008 #8


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    You have two equations to solve for x and y. Solve the equations simultaneously.
  10. Nov 6, 2011 #9
    Do you mean set up the two equations equal to each other, then plug in -x for all y's then solve for x then plug that back into the equation to get the y?
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