# Equation of tangent line

• April 30th 2012, 12:34 PM
Foxlion
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 $f'(x)$?????
• April 30th 2012, 12:38 PM
earboth
Re: Equation of tangent line
Quote:

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:

$y-\frac12 = -\frac18(x-4)~\implies~\boxed{y = -\frac18x+1}$
• April 30th 2012, 12:39 PM
MathoMan
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 $l$.
• April 30th 2012, 12:40 PM
MathoMan
Re: Equation of tangent line
Earboth beat me to it. :D
• April 30th 2012, 12:43 PM
Foxlion
Re: Equation of tangent line
Quote:

Originally Posted by Foxlion
$y = -\frac{1}{8}x + ?$

never mind
• April 30th 2012, 12:45 PM
HallsofIvy
Re: Equation of tangent line
Are you saying that earboth and mathoman were just wasting their time responding to your post?
• April 30th 2012, 01:19 PM
Foxlion
Re: Equation of tangent line
Quote:

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 $?$