# How many Tangent lines go to a certain point.

Hypnos_16

## Homework Statement

How many tangent lines to the curve y = x / (x + 1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

n/a

## The Attempt at a Solution

I'm completely drawing a blank on this one. Help.

Homework Helper
Let's first simply imagine a tangent line to the curve passing through the point (1,2).
What would the equation of this tangent be?

Well we know a point, and we don't know what the gradient is but we do know it is tangent to the curve so this would point towards the derivative equation, yes?

So we have, $$y-2=m(x-1)$$ where $$m=\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{x+1}\right)$$

And this line equation touches the curve, so what can we do here?

Hypnos_16
So we have, $$y-2=m(x-1)$$ where $$m=\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{x+1}\right)$$

And this line equation touches the curve, so what can we do here?

Find the slope of the line, and then follow it outward to see at what other points it would touch the graph?

Homework Helper
Find the slope of the line, and then follow it outward to see at what other points it would touch the graph?

Trial and error is one way, but a very bad way. Let's do this more rigorously

You already have the equation of the tangent line(s), $$y=m(x-1)+2$$ and the equation of the curve $$y=\frac{x}{x+1}$$, and we know that the line touches the curve. How can we find the point where they touch by mathematical means?

If you can't quite figure it out, it's probably that you're just being overwhelmed by the problem at hand. As a simpler example, it's equivalent to finding the points of intersection between the equations $$y=x+1$$ and $$y=2x$$

Hypnos_16
you could make them equal one another. Since they both have y = (something) then solve for the variable. Or that's what i'd try in your example there. But i don't know how to ramp that up to make it work on a bigger scale, like the example i have.

Homework Helper
That's it.

So we just have to solve $$m(x-1)+2=\frac{x}{x+1}$$ for x, where $$m=\frac{dy}{dx}$$. This will give you the points where the line and curve intersect, which means you'll have the x values of where the tangent(s) to the curve passing through (1,2) touch the curve.

Hypnos_16
Alright, so then i would have to first get the slope of the line though wouldn't i? how can i do that if i only have one point?

Homework Helper
Alright, so then i would have to first get the slope of the line though wouldn't i? how can i do that if i only have one point?

Well we know the gradient is going to be the derivative at the point x that we solve for, so we can just find the derivative and replace the variable m for it.

Hypnos_16
So i got a value of x to be -2 - 1√3 and -2 + 1√3. I'm thinking something has gone wrong, cause that would be impossible to try and work out.

Homework Helper
What do you mean it would be impossible to try and work out?

So the x values are $$x=-2\pm \sqrt{3}$$, now find the corresponding y values to complete the answer. If you want to check to be sure you have the right answer, find the derivative at those points, then test to see if the line going through that point with such a gradient passes through (1,2).

But I'll save you the effort and tell you now that it's right.

Hypnos_16
haha thanks. I just had a moment where i forgot these can also be decimal values. As i read that too i thought, this problem doesn't seem worth all this effort. But thanks for checking them for me. I appreciate it man. Thanks for all your help again.