# Find the equation of the line tangent to two parabolas

1. Feb 13, 2013

### flemonster

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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.

2. Feb 13, 2013

### iRaid

3. Feb 13, 2013

### haruspex

You do not appear to have used the fact that the points (x1, y1), (x2, y2) lie on the respective parabolas.

4. Feb 13, 2013

### flemonster

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?

5. Feb 13, 2013

### flemonster

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!

6. Feb 13, 2013

### haruspex

I mean that e.g. x1, y1 satisfy the equation of the first parabola.