# Finding tangent lines to a curve

Find the equations of all tangent lines to the curve $y=x/(x+1)$ that intersect the point $(1,2)$. Note that the point $(1,2)$ does not lie on the curve. Please simplify your final answer as much as possible.

-I am so lost in trying to solve this question. I found the derivative of $f(x)$ which is $1/(x+1)^2$ which is equal to the slope. Then I wrote the equation in the $y=mx +c$ format to solve for an equation. I got the equation $y=[1/(x+1)^2]x +1.75$

I have a feeling this is incorrect and do not know what to do next. Am I approaching this question right?

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Using all caps is considered shouting on the internet, and is quite rude. Please don't do it again. –  Zev Chonoles Sep 25 '13 at 1:55
Please don't shout on the internet. I've edited your title to be germane to the question. –  T. Bongers Sep 25 '13 at 1:56

The derivative is correct.

As stated in the execise, the point $(1,2)$ is not on the curve; so you need to find such a point $(x_0,y_0)$ which a) lies on the curve; b) the tangent line to the curve in this point passes through $(1,2)$. Are these hints sufficient?

Edit

Our line is described by the equation $y=kx+c$. It is tangent to the curve int he point $(x_0,y_0)$, thus $y_0=kx_0+c$; in addition, the derivatives of the line and of the curve must coincide in this point, i.e. $k=\frac{1}{(x_0+1)^2}$. Finally, the line passes though $(1,2)$, hence $2= k+c$.

Thus, we obtain the equations: $$y_0=kx_0+c\\y_0=x_0/(x_0+1)\\k=\frac{1}{(x_0+1)^2}\\2= k+c$$

Can you solve these equations to obtain all possible pairs $(k,c)$?

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So I need to find points which lie on the curve but not the tangent line? So i plug in x=1 in to the f(x) equation to fine y? Sorry I am still confused –  Lisa Paima Sep 25 '13 at 2:01
@LisaPaima see edit. –  TZakrevskiy Sep 25 '13 at 2:08
So now do i use two of the equations to form simultaneous equations then solve for k and c? –  Lisa Paima Sep 25 '13 at 2:17
@LisaPaima As one of possibilities, yes. Another strategy would be to express $k$ and $c$ in terms of $x_0$ and $y_0$ and then solve for $x_0$, $y_0$. I suggest you try the latter. –  TZakrevskiy Sep 25 '13 at 2:20
Alright, I am going to try it out thank you so much! –  Lisa Paima Sep 25 '13 at 2:22
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A slightly different way of attacking the problem:

Let $(X,Y)$ be a point on the given curve, and also on a straight line through $(1,2)$

Since $(X,Y)$ is on the curve, we have Equation #$1$ $$Y=\frac{X}{X+1}$$

Using your derivative, the slope of the curve at $(X,Y)$ is:$$\text{Curve Slope} =\frac{1}{(X+1)^2}$$The slope of a straight line through $(X,Y)$ and $(1,2)$, $\text{m}$, is given by:$$\text{m}=\frac{Y-2}{X-1}$$Setting the curve slope equal to the slope of the straight line. we get Equation #$2$:$$\frac{1}{(X+1)^2}=\frac{Y-2}{X-1}$$If we use Equation#$1$ to eliminate $Y$ from Equation #$2$ we get an equation in $X$ that can be re-arranged into a quadratic.
EDIT:$$\frac{1}{(X+1)^2}=\frac{\frac{X}{X+1}-2}{X-1}=\frac{X-2X-2}{(X+1)(X-1)}$$ $$\frac{1}{(X+1)}=\frac{-X-2}{(X-1)}$$

Solve the quadratic, sub back into #$1$ to get $Y$. Each $(X,Y)$ combined with $(1,2)$ gives two points on a straight line, and thus the equation of the two tangent lines.

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thank you so much, this was a lot of help –  Lisa Paima Sep 25 '13 at 3:10
When y is substituted in to the equation it's making it really hard to form a quadratic equation. Im trying to cross multiply them and I get x^3, it seems incorrect to me. How can I cross multiply this out to get a quadratic so I could solve for 'x' –  Lisa Paima Sep 25 '13 at 3:41
When you substitute for $Y$, you get $X+1$ on the denominator of the RHS. Since $X$ cannot be $-1$, it is safe to cancel out one factor of $X+1$ from both sides. So, no cubic... –  User58220 Sep 25 '13 at 7:04

The general form of a line which intersects the point $\left(1,2\right)$ is given by $y = 2 + a\left(x - 1\right)$ where $a$ is the line slope. The line and curve intersection is determined by

$$2 + a\left(x - 1\right) = x/\left(x + 1\right)$$

Since the line slope $a$ must be equal to the curve slope at the intersection point, we have

$$a = {{\rm d} \over {\rm d}x}\left(x \over x + 1\right) = {1 \over \left(x + 1\right)^{2}}$$ In solving those equations we find two values of $a\ \left(7 \mp 4 \sqrt{3\,}\right)$ which yield two lines:

$$\begin{array}{|lcr|}\hline\\ \quad\color{#ff0000}{\large y} & \color{#000000}{\large\ =\ } & \color{#ff0000}{\large% 2 + \left(7 - 4\sqrt{3\,}\,\right)\left(x - 1\right)\quad} \\[2mm] \quad\color{#ff0000}{\large y} & \color{#000000}{\large\ =\ } & \color{#ff0000}{\large% 2 + \left(7 + 4\sqrt{3\,}\,\right)\left(x - 1\right)\quad} \\ \\ \hline \end{array}$$

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Where did 'a' come from? I do not understand where the ax + 2 -a comes from..?! –  Lisa Paima Sep 25 '13 at 2:50
I have no idea how you started off the question –  Lisa Paima Sep 25 '13 at 2:58
Since $\left(1,2\right)$ belongs to the line, the general form is $y = 2 + a\left(x - 1\right)$: When $x = 1$, we get $y = 2$. For the time being, we don't know anything more about $a$. If the line intersects the curve as a tangent one, there is just one point of intersection. Then, we determine that point and require that we have just one point. The second degree equation must have one solution. That condition yields an equation for $a$ which is what we need. –  Felix Marin Sep 25 '13 at 3:22
How did you get to the final answer? what was the equation that you solved to find both values for a –  Lisa Paima Sep 29 '13 at 23:57