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Limit word problem, find area of triangle

  1. Oct 16, 2006 #1
    1. the tangent to the curve xy=4 at a point P in the first quadrent meets the x-axis at A and the y-axsis at B. Prove that the area of triangle AOB, where O is the origin, is independant of the position P.

    y=4/x
    Find thd slope of the Tangent
    lim f(a+h)-f(a) /h
    lim 4/a+h - 4/a /h
    lim 4a - 4a -4h/ a(a+h) /h
    lim -4h / h(a(a+h))
    lim -4/a(a+h)
    -4/h
    Quardinents of point P

    p(a,4/a)

    the equation for the tangent

    y-y1=m(x-x1)
    y - 4/a = -4/a^2 (x - a)
    y - 4/a = (-4/a^2)x + 4a/a^3
    y = (-4/a^2)x + 4a/a^3 + 4/a
    y = (-4/a^2)x + 8a/a^3

    but the answer is y = (-4/a^2)x + 8/a can someone show me the steps to solving this?

    I can do the rest I just need help with this. Thanks
     
    Last edited: Oct 16, 2006
  2. jcsd
  3. Oct 16, 2006 #2
    [tex] y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{3}} [/tex] should be [tex] y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{2}} [/tex].

    So you have [tex] y - \frac{4}{a} = -\frac{4}{a^{2}}(x-a) [/tex].

    [tex] y - \frac{4}{a} = -\frac{4x}{a^{2}} + \frac{4a}{a^{2}} [/tex]

    [tex] y = -\frac{4x}{a^{2}} + \frac{4}{a} + \frac{4}{a} [/tex].

    [tex] y = -\frac{4x}{a^{2}} + \frac{8}{a} [/tex].
     
    Last edited: Oct 16, 2006
  4. Oct 16, 2006 #3
    Thanks courtrigrad
     
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