Find the equation of the line tangent to two parabolas

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Homework Statement


Given the two parabolas: f(x) = x^2 - 2x + 2 and g(x) = -x^2 - 2x - 2. Find the equation of a line that is tangent to both curves.

Homework Equations



The given parabolas, equation for a line y = mx + b, and the derivatives of the two parabolas 2x - 2 and -2x - 2

The Attempt at a Solution



The line tangent to the two parabolas will pass through the points

(x_1 , y_1)

for the parabola f(x) and

(x_2 , y_2)

for the parabola g(x)

so the equations for the two lines will be,

for f' y_1 = (2x_1 - 2)x_1 +b

and

g' y_2 = (-2x_2 - 2)x_2 + b.

Since the slopes of both lines will be the same I thought that setting the two slopes equal might get me started so I wrote

2x_1 - 2 = -2x_2 - 2

which gave me \frac{x_1}{x_2} = -1.

I rearranged the two linear equations and set them equal:

y_1 - (2x_1 - 2)x_1 = y_2 - (-2x_2 - 2)x_2

but that got me absolutely nowhere. I got the whole thing down to

x^2 _1 + x^2 _2 = \frac{y_1 - y_2}{2}

but that doesn't help.

I know I need to limit my variables and try to get the whole thing in terms of one variable but I'm at a loss as to how to make that happen. Every time I substitute and simplify I get either one or negative one which tells me nothing. I can't figure out how relate the equations and simplify. Any help is appreciated.
 
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http://www.math.washington.edu/~aloveles/Math124Fall2012/ConroytangentLines01.pdf

I don't know how to do these problems, but that looks exactly how I would go about it. Sorry I can't give any additional help.
 
You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.
 
haruspex said:
You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.

I'm not sure what you mean. I accounted for them being on different parabolas with the subscripts but is there something in their being on different parabolas that I missed?
 
And I just figured it out. That tutorial from UW did the trick which is kind of funny seeing that's where I did my undergrad. Thanks iRaid!
 
flemonster said:
I'm not sure what you mean.
I mean that e.g. x1, y1 satisfy the equation of the first parabola.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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