# Thread: Equation of tangent line

1. ## Equation of tangent line

Find equation of line tangent to the hyperbola $\frac{2}{x}$ at the point $(4,1/2)$

I can't see where the y-intercept comes from... $f' = -2x^{-2} \Rightarrow f'(4) = -\frac{1}{8}$ so the equation is $y = -\frac{1}{8}x + ?$

I don't where to go from finding the slope, do I just keep plugging in points for $f'(x)$?????

2. ## Re: Equation of tangent line

Originally Posted by Foxlion
Find equation of line tangent to the hyperbola $\frac{2}{x}$ at the point $(4,1/2)$

I can't see where the y-intercept comes from... $f' = -2x^{-2} \Rightarrow f'(4) = -\frac{1}{8}$ so the equation is $y = -\frac{1}{8}x + ?$

I don't where to go from finding the slope, do I just keep plugging in points for $f'(x)$?????
You have the coordinates of the tangent point and the slope of the tangent line. Use slope-point-equation of a straight line:

$y-\frac12 = -\frac18(x-4)~\implies~\boxed{y = -\frac18x+1}$

3. ## Re: Equation of tangent line

$y=-\frac{1}{8}x+l$ and $l$ can be found using the condition that the point $\left(4,\frac{1}{2}\right)$ belongs to that tangent line. Use the coordinates of that point instead of x and y in and you will get $l$.

4. ## Re: Equation of tangent line

Earboth beat me to it.

5. ## Re: Equation of tangent line

Originally Posted by Foxlion
$y = -\frac{1}{8}x + ?$
never mind

6. ## Re: Equation of tangent line

Are you saying that earboth and mathoman were just wasting their time responding to your post?

7. ## Re: Equation of tangent line

Originally Posted by HallsofIvy
Are you saying that earboth and mathoman were just wasting their time responding to your post?
no, I posted that before seeing their replies. I looked at what I had posted and saw that I simply had to plug in an x and a y and solve for the $?$