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,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.
Hello,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
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.