# How to prove this hyperbola question

• Feb 19th 2006, 08:12 AM
totalnewbie
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.
• Feb 19th 2006, 10:41 AM
ThePerfectHacker
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.
• Feb 19th 2006, 11:36 AM
earboth
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
• Feb 19th 2006, 02:59 PM
ThePerfectHacker
My fault, I solved the problem by assuming the triangle was right. Thus, I solved a completely different problem, sorry. :o
• Feb 19th 2006, 04:23 PM
ThePerfectHacker
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.
• Feb 20th 2006, 06:31 AM
totalnewbie
Thank you very much !!!
• Feb 20th 2006, 11:54 AM
ThePerfectHacker
Welcome, I really find this problem elegant. Where did you find it?
• Feb 20th 2006, 12:45 PM
totalnewbie
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.