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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- Feb 19th 2006, 07:12 AMtotalnewbieHow 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.

- Feb 19th 2006, 09:41 AMThePerfectHackerQuote:

Originally Posted by**totalnewbie**

$\displaystyle x'>0$.

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. - Feb 19th 2006, 10:36 AMearbothQuote:

Originally Posted by**ThePerfectHacker**

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 - Feb 19th 2006, 01:59 PMThePerfectHacker
My fault, I solved the problem by assuming the triangle was right. Thus, I solved a completely different problem, sorry. :o

- Feb 19th 2006, 03:23 PMThePerfectHacker
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. - Feb 20th 2006, 05:31 AMtotalnewbie
Thank you very much !!!

- Feb 20th 2006, 10:54 AMThePerfectHacker
Welcome, I really find this problem elegant. Where did you find it?

- Feb 20th 2006, 11:45 AMtotalnewbie
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.