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Thread: 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 $\displaystyle x'$ be on $\displaystyle 4/x$. For simplicity sake assume,
    $\displaystyle x'>0$.

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

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

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

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

    Similarily it is true when $\displaystyle x'<0$. 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 $\displaystyle x'$ be on $\displaystyle 4/x$. For simplicity sake assume,
    $\displaystyle x'>0$.
    ...
    Similarily it is true when $\displaystyle x'<0$. 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 $\displaystyle 2x'-x'=x'$
    And its height to be $\displaystyle 4/x'$. Thus, the area must be $\displaystyle \frac{1}{2}\frac{4}{x'}x'=2$ a constant for whatever value of $\displaystyle x'$.

    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 $\displaystyle \frac{8}{x'}$.

    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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    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 $\displaystyle x'$ in my other post is,
    $\displaystyle y-\frac{4}{x'}=-\frac{4}{(x')^2}{x-x'}$.
    Now, the x-axis and y-axis are formed when y is zero and x is zero respectively.
    Thus,
    When y=0
    Then,
    $\displaystyle -\frac{4}{x'}=-\frac{4}{(x')^2}({x-x'})$
    Thus, $\displaystyle x=2x'$
    When x=0
    Then,
    $\displaystyle y-\frac{4}{x'}=-\frac{4}{(x')^2}{(0-x')}$
    Thus,
    $\displaystyle y=\frac{8}{x'}$
    Now, the area is,
    $\displaystyle \frac{1}{2}\frac{8}{x'}\frac{2x'}{1}=8$
    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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