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Here is the question and i dont really understand

Point $(a,b)$ is on the function $f(x)=\frac{2}{x}$ $x>0$. Show that the area of the triangle formed by the tangent line at $(a,b)$ , the $x$ axis and $y$ axis is equals to $4$.

What is the question asking?

I used the first principle to find the derivative $f'(x)=\frac{-2}{x^2}$

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    First, can you find the equation of the tangent line through $(a,b) $? – 2017-02-21
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    we need to find the slope , but i dont know how – 2017-02-21

2 Answers 2

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  1. Find $b$ in terms of $a$ using $f(x)=2/x$. This gives you the point $(a,b)$ in terms of $a$.
  2. Find the slope of the tangent line at $(a,b)$ using the derivative of $f$, in terms of $a$.
  3. Find the $y$-intercept of the line using the point and the slope.
  4. Use the equation of the line to get the $x$-intercept.
  5. You now have two sides of a right triangle using the $x$ and $y$ intercepts. Use the area formula for a triangle to show that the area must be 4.
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    How do I find b if I dont know which point is the tangent line that will create a triangle that have aera of 4? – 2017-02-21
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    The point $(a,b)$ tells you that $x=a$ and $y=f(a)=2/a=b$ so $2/a=b$. – 2017-02-21
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    a and b can be any point on the graph so how do i know which one is it to find the slope of the tengant line? – 2017-02-21
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    The main underlying point is that it doesn't matter. No matter what value you choose for $a$ the area is always the same. – 2017-02-21
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    Thanks, tried every point and find the slope of the tangent and the area stilll the same. Thanks for the help – 2017-02-21
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    @PaulSundheim I have carefully computed the area several times without it coming out to 4. What am I doing wrong? Here are my calculations: the derivative is $f'(x) = \frac{-2}{x^2}$, implying the slope at $(a,\frac{1}{a})$ is $\frac{-2}{a^2}$. Hence, the tangent line at $(a,\frac{1}{a})$ is $y = \frac{-2x}{a^2} + \frac{3}{a}$. The $x$ and $y$ intercepts are $(\frac{3}{2}a,0)$ and $(0,\frac{3}{a})$. Hence, the area is $A = \frac{1}{2} \cdot \frac{3a}{2} \cdot \frac{3}{a} = \frac{9}{4}$. I know I am terrible at calculations, but... – 2017-02-23
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    You are using $(a,1/a)$ when you should be using $(a,2/a)$. I get the $y$-intercept to be $4/a$ and the $x$-intercept as $2a$. – 2017-02-23
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The questioner asked for help understanding the question itself, and here a picture is worth $10^3$ words:

enter image description here

The goal is then to find the value of $a$ (and then $b$) such that the area of the triangle defined by the tangent and the axes has the desired value. The figure above is to illustrate a candidate solution. The goal is to use simple algebra to find the actual solution values. Paul Sundheim's approach is straightforward indeed.

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    How do you know that the tangent line is at the red point and not (1,2) or (2,1)? – 2017-02-21
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    @David The goal is to show that the area of the triangle is always 4 (i.e., independently of the choice of $(a,b)$). – 2017-02-21
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    @Secret The tangent line, according to the problem, goes through $(a,b)$, which is a point anywhere on the curve. – 2017-02-21