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  1. Jan 26, 2005 #1
    Let y=f(x) be the continuous function that satifies the equation x^4-5x^2y^2+4y^4=0 and whose graph contains the points (2,1) and (-2,-2). Let 'L' be the tangent to the graph of f(x) at x=2.

    A) Find an expression for y'
    B) Write an equation for the line 'L'
    C) Give the coordinates of a point that is on the graph of f(x) but not the line 'L'

    I am assuming part A means to take a derivative (which I did) and got
    (-4x^3+10xy^3)/(-5x^2 2y+16y^3) I don't know if this is right but now I am stuck...any help in the right direction would be great! Thanks!
  2. jcsd
  3. Jan 26, 2005 #2
    ok I think your function is [tex] y = f(x) = x^4 - 5x^2y^2 + 4y^2 = 0 [/tex]. So for art a you want to differentiate implicitly with respect to x. For the second part use point slope for and substitute [tex] x = 2 [/tex]. For the third part you know what point is on both graphs? SO give a point which is on [tex] f(x) [/tex] but not on [tex] L [/tex]. For the derivative I got [tex] \frac {4x^3 + 10xy^2}{10x^2y + 8y} [/tex]
    Last edited: Jan 26, 2005
  4. Jan 26, 2005 #3


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    Is your function:
    [tex] F(x,y)=x^{4}-5x^{2}y^{2}+4y^{4} [/tex]

    Use the theorem of implicit functions for point "a"...


    P.S.Then the other 2 points are easy to solve.
  5. Jan 26, 2005 #4
    dextercoiby has the correct function...any more help would be great...
  6. Jan 26, 2005 #5


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    Did u compute the
    [tex] \frac{dy}{dx} [/tex]

    in the points in which the explicitation is possible...?

  7. Jan 26, 2005 #6
    yes and i got that answer...im just not sure where to go with it now...
  8. Jan 26, 2005 #7
    you know L' is tangent to x = 2. The coordinates are (2,1). Now use point slope form to get the equation.

    [itex] y-y_1 = m(x-x_1) [/itex]
  9. Jan 27, 2005 #8


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    "m" is the slope in the point (2,1).So if you computed the derivative,the slope is immediate...Then use the formula written by the previous poster to solve point "b",too...

    As for point "c",i think it's not that difficult.

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