Re: Equation of tangent line

Quote:

Originally Posted by

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

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

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

You have the coordinates of the tangent point and the slope of the tangent line. Use slope-point-equation of a straight line:

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

Re: Equation of tangent line

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

Re: Equation of tangent line

Earboth beat me to it. :D

Re: Equation of tangent line

Quote:

Originally Posted by

**Foxlion** $\displaystyle y = -\frac{1}{8}x + ?$

never mind

Re: Equation of tangent line

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

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 $\displaystyle ?$