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I have no idea how to do this problem at all.

Find equation of the tangent lines to the curve $$y = \frac {x-1}{x+1}$$ that are parallel to the line $x-2y = 2$

I found the derivative of $y = \frac {x-1}{x+1}$ to be $\frac 2{x^2 + 2x +1}$ and the other line I rewrote as $\frac{-2+x}{2}$

From here I attempted to set them equal to each other and I got

$6 = x^3 - 3x$ which is incredibly difficult to solve and is probably not the answer. Where did I go wrong?

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When lines are parallel they have the same slope. What's the slope of $x-2y=2$? What's the relation of derivative to slopes and tangent lines? – Gerry Myerson May 10 '12 at 1:50
I get that the slope is 1/2. A derivative is a slope of a tangent line. – user138246 May 10 '12 at 1:57
up vote 3 down vote accepted

The line $x-2y=2$ has slope $\frac12$, so you want to set the derivative of $\frac{x-1}{x+1}$ equal to $\frac12$ and solve for $x$:


This would actually be just a little easier if you hadn't multiplied out the denominator; then you'd have $$\frac2{(x+1)^2}=\frac12\;,\tag{1}$$ so $(x+1)^2=4$, $x+1=\pm 2$, ...

Added: Let's recap what's going on here. You want some lines that are to be parallel to the line $x-2y=2$. That line has slope $1/2$, so your lines have to have slope $1/2$: that's how you know that they're parallel to the line $x-2y=2$. (If they had a different slope, they'd eventually cross it somewhere.) The derivative $\frac2{(x+1)^2}$ gives the slope of the tangent at any point on the curve $y=\frac{x-1}{x+1}$; we want to know where that slope is $1/2$, so we set the derivative equal to $1/2$ in $(1)$ and solve for $x$. You found that $x=-3$ or $x=1$, so there are two points on the curve where the tangent is parallel to the line $x-2y=2$; one is $(-3,2)$, and the other is $(1,0)$. Now you just find the tangent lines to $y=\frac{x-1}{x+1}$ at those two points.

For the first, for instance, you have $y-2=\frac12\big(x-(-3)\big)=\frac12(x+3)=\frac12x+\frac32$, so $y=\frac12x+\frac72$; I'll leave the other to you.

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I am not sure what to do I get x= -3 or 1 but that doesn't seem to help, I am not sure where these numbers go. – user138246 May 10 '12 at 2:03
@Jordan: They are the $x$-coordinates of the points on the curve $y=\frac{x-1}{x+1}$ where the slope is $\frac12$. Now you need to find the $y$-coordinates of those points and the tangent lines to $y=\frac{x-1}{x+1}$ at those points. Those tangent lines will be parallel to each other and to the given straight line, since they both have slope $\frac12$. – Brian M. Scott May 10 '12 at 2:08
I think I got the answer then, for the x points -3 and 1 the lines will be parral at the points (1, 0) (-3, -2) and (1, -1) (-3, 05/2) – user138246 May 10 '12 at 2:24
@Jordan: Have you found the points on $y=\frac{x-1}{x+1}$ where $x$ is $-3$ and $1$? Then just find the equations of the tangent lines at those points. Ignore $x-2y=2$: we've no more use for it. When you're done, you'll find that your two tangent lines both have slope $1/2$; and that automatically makes them parallel to $x-2y=2$, since its slope is also $1/2$. – Brian M. Scott May 10 '12 at 2:27
@Jordan: The first one can't be right: its slope isn't $1/2$. Hold on: I'm expanding my answer. – Brian M. Scott May 10 '12 at 2:34

From your equation of the line, the slope is 1/2. This is what you set the derivative equal to, and then solve for x. From x, using the original function, get y.

So now you have the point $(x, y)$ and the slope; using the point-slope (duh) form of a line, get the equation of the tangent line.

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Hint: You have somewhat randomly set two things equal to each other. What you currently have is finding the intersection between the derivative and the line. What you want is to set the derivative equal to the slope of that line, in order that they are parallel.

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A quite minor point:

I would rather call the $x$-coordinate of the point(s) of tangency by some name other than $x$, like $a$.

Since $\frac{dy}{dx}=\frac{2}{(x+1)^2}$, we conclude that the tangent line at $x=a$ has slope $\frac{2}{(a+1)^2}$.

Then we need to set $\frac{2}{(a+1)^2}=\frac{1}{2}$, and solve for $a$. After that, the computation is the familiar one.

A similar issue comes up when we ask, for example, for the equations of all tangent lines to the curve $y=x^2$ that pass through the point $(1,-3)$. Note that $(1,-3)$ is not on the curve. If we let the point of tangency be $(a,a^2)$, then the equation of the tangent line is $y-a^2=2a(x-a)$. Substituting $1$ for $x$ and $-3$ for $y$, we obtain a quadratic equation for $a$. Solve, and we are essentially finished. More than once I have seen a student go astray by using $x$ for the $x$-coordinate of the point of tangency.

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I do not understand the $(a,a^2)$ part. – user138246 May 10 '12 at 12:56
Don't worry about it, the last paragraph is not about the question that you asked, it is about a harder question that you have not worked on. – André Nicolas May 10 '12 at 13:02

The equations of tangent lines that are parallel is y-y1 = (1/2)(x-1) for all y1 in real numbers.

Solution: The slope of given curve is dy/dx = 2/(x+1)^2 We have to find equations of tangent lines that are parallel that means If we take any two tangent lines at (x1,y1) and at (x2,y2) that are parellal then slopes of those equations should be equal.

That means 2/(x1 + 1)^2  = 2/(x2 + 1)^2

Solve it then you will get x1 = x2 = 1

So the tangent lines at (1,y1) that are parallel for all y1 in real numbers, are y-y1 = (1/2)(x-1)

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