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**

.

Now the tangent line to point is the derivative at that point. Thus, is the slope of the derivative. Now find the equation of the tangent.

The equation of line having slope and point is (by point slope formula is):

Now the x-intercept is when . Thus,

Thus,

Thus,

Thus, it intercepts the x-axis at .

This would imply that the base of the triangle must be

And its height to be . Thus, the area must be a constant for whatever value of .

Similarily it is true when . 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

And its height to be . Thus, the area must be a constant for whatever value of .

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 .

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 in my other post is,

.

Now, the x-axis and y-axis are formed when y is zero and x is zero respectively.

Thus,

**When y=0**

Then,

Thus,

**When x=0**

Then,

Thus,

Now, the area is,

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.