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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.

This is what I have done so far. I don't know if it is correct.

1/(x+1)= -x-2/(x-1) [cross-multiplied] to get, x-1=-x^2 -3x -2

I made it equal to zero to get a quadratic equation of x^2 + 4x + 1 = 0 to solve for x I got x = -2 +- root 3

I dont know what to do next.

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Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:

$$y'(x)=\frac{1}{(1+x)^2}$$

so the slope of the tangent line at our point $(a,b)$ is

$$y'(a)=\frac{1}{(1+a)^2}$$

The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is

$$y-b=\frac{1}{(1+a)^2}(x-a)$$

We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:

$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$

This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation

$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$

to obtain the equations of the desired tangent lines.

Here is a helpful picture plotted with Wolfram|Alpha: enter image description here

Notice where the tangent lines intersect!

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  • $\begingroup$ So the quadratic that I found with the solutions for a is correct? $\endgroup$
    – Lisa Paima
    Sep 30 '13 at 1:20
  • $\begingroup$ @LisaPaima: Yes. Although I'm not exactly sure how you found it, but it is the correct answer. $\endgroup$
    – Jared
    Sep 30 '13 at 1:22
  • $\begingroup$ Oh okay good, thank you. One more question; when I am substituting a into the formula to give me the equations for the tangent lines, do I have to substitute any values for x? $\endgroup$
    – Lisa Paima
    Sep 30 '13 at 1:25
  • $\begingroup$ @LisaPaima: Great question. The equation of a line represents infinitely many points, and thus when writing the equation of a line, we should keep $x$ as a variable. Plugging in a constant value for $x$ will specify a particular point on the line, and thus not represent the entire line. $\endgroup$
    – Jared
    Sep 30 '13 at 1:27
  • $\begingroup$ I substituted 'a' into the equation, and for one of my a values I got: -0.25x + 1.366025404 .. It's not meant to be in decimals. I'm not sure if this is correct.. $\endgroup$
    – Lisa Paima
    Sep 30 '13 at 1:31

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