# Thread: Equations of lines tangent to a curve

1. ## Equations of lines tangent to a curve

I'm stuck on the following problem:

"Find the equations of both lines through the point (1,2) that are tangent to the curve y=x/(x+1)"

So far, I have that y'=1/(x+1)^2. I tried setting that equal to the slope (2-(x/(x+1)))/(1-x) but it isn't working out. Not sure where to go from here, any help would be much appreciated

2. Originally Posted by maryanna91
I'm stuck on the following problem:

"Find the equations of both lines through the point (1,2) that are tangent to the curve y=x/(x+1)"

So far, I have that y'=1/(x+1)^2. I tried setting that equal to the slope (2-(x/(x+1)))/(1-x) but it isn't working out. Not sure where to go from here, any help would be much appreciated

What you wanted to do is right: to compare the slope between (1,2) and a generic point

$\left(x,\frac{x}{x+1}\right)$ on the function's graph, and the slope of the tangent

line at this point given by the derivative:

$\displaystyle{\frac{2-\frac{x}{x+1}}{1-x}=\frac{1}{(x+1)^2}\Longrightarrow 2(x+1)^2-x(x+1)=1-x\Longrightarrow x^2+4x+1=0\Longrightarrow$

$x_{1,2}=-2\pm\sqrt{3}$ , and thus the slopes are $\displaystyle{m_{1,2}=\frac{1}{4\pm 2\sqrt{3}}=1\pm \frac{\sqrt{3}}{2}$

Now use the well-known formula for the equation of a line through a point given its

slope to find both equations.

Tonio

3. Looks like my algebra was just off, thanks a lot for your help!