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Math Help - How to prove this hyperbola question

  1. #1
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    How to prove this hyperbola question

    How to prove that hyperbola y=4/x determines a triangle which area is 8 when I draw tangent point at whatever point on this hyperbola and when I consider that x-axis, y-axis and tangent line form triangle.
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  2. #2
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    Quote Originally Posted by totalnewbie
    How to prove that hyperbola y=4/x determines a triangle which area is 8 when I draw tangent point at whatever point on this hyperbola and when I consider that x-axis, y-axis and tangent line form triangle.
    Let x' be on 4/x. For simplicity sake assume,
    x'>0.

    Now the tangent line to point (x,y)=(x,4/x) is the derivative at that point. Thus, -4x^{-2} is the slope of the derivative. Now find the equation of the tangent.

    The equation of line having slope -4(x')^{-2} and point (x,4/x) is (by point slope formula is):
    y-\frac{4}{x'}=-\frac{4}{(x')^2}(x-x')

    Now the x-intercept is when y=0. Thus,
    -\frac{4}{x'}=-\frac{4}{(x')^2}(x-x')
    Thus,
    x'=x-x'
    Thus,
    2x'=x
    Thus, it intercepts the x-axis at 2x'.

    This would imply that the base of the triangle must be 2x'-x'=x'
    And its height to be 4/x'. Thus, the area must be \frac{1}{2}\frac{4}{x'}x'=2 a constant for whatever value of x'.

    Similarily it is true when x'<0. For one thing this hyperbola is symettric.

    Q.E.D.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Let x' be on 4/x. For simplicity sake assume,
    x'>0.
    ...
    Similarily it is true when x'<0. For one thing this hyperbola is symettric.

    Q.E.D.
    Hello,

    I'm a little bit puzzled: You wrote: This would imply that the base of the triangle must be 2x'-x'=x'
    And its height to be 4/x'. Thus, the area must be \frac{1}{2}\frac{4}{x'}x'=2 a constant for whatever value of x'.

    1. As I understand the problem the base of the right triangle must be 2x'.
    2. You get the interception with the y-axis if x=0. Then the height of the triangle is  \frac{8}{x'}.

    If you take these results the value of the area will be 8 as totalnewbie has written.

    Greetings

    EB
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  4. #4
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    My fault, I solved the problem by assuming the triangle was right. Thus, I solved a completely different problem, sorry.
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  5. #5
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    Okay let me start again the triangle is formed by the intersecting line between the y-axis the x-axis.
    Since the equation of the of tangent at point x' in my other post is,
    y-\frac{4}{x'}=-\frac{4}{(x')^2}{x-x'}.
    Now, the x-axis and y-axis are formed when y is zero and x is zero respectively.
    Thus,
    When y=0
    Then,
    -\frac{4}{x'}=-\frac{4}{(x')^2}({x-x'})
    Thus, x=2x'
    When x=0
    Then,
    y-\frac{4}{x'}=-\frac{4}{(x')^2}{(0-x')}
    Thus,
    y=\frac{8}{x'}
    Now, the area is,
    \frac{1}{2}\frac{8}{x'}\frac{2x'}{1}=8
    Q.E.D.
    Here is diagram below.
    Attached Thumbnails Attached Thumbnails How to prove this hyperbola question-picture6.gif  
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    Thank you very much !!!
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  7. #7
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    Welcome, I really find this problem elegant. Where did you find it?
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  8. #8
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    I am preparing for my math exam and that is why I decided to borrow one math book which contains excercises for people who want to pass math exam. There are many difficult excercises.
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