Expert: Paul Klarreich - 10/16/2007

Question
Could you help me with this?  Graph the two parabolas y=x^2 and y=-x^2+2x-5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.

Answer
Questioner:   Bryce
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  Could you help me with this?  Graph the two parabolas y=x^2 and y=-x^2+2x-5 in the same coordinate plane. Find the equations of the two lines simultaneously tangent to both parabolas.
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Hi, Jeff,
I mean,

Hi, Bryce,

[Sorry -- got you mixed up with another student who asked this question last month.  You really should look at the question archives; there is lots of really cute stuff there.]

It isn't easy -- you have to do some detective work and some algebra.  (and a teensy bit of calculus, too)

Here are your two parabola:

I.  y = x^2

II. y = - x^2 + 2x - 5

Some line will have these properties:(yes, I know, there are supposed to be two of them)

1. It intersects both graphs.
2. It is tangent at those points.

Let:

a = the value of x where the line touches graph I.
b = the value of x where the line touches graph II.

Let's work on a:

The line passes through the point  (a,a^2).
At that point,  y' = 2x,  so  m = 2a.

Let's work on b:

The line passes through the point  (b,-b^2+2b-5).
At that point,  y' = -2x + 2,  so  m = -2b + 2.

But this is the same line, so we can also compute the slope of the line by ye-olde-slope-formula:
   y2 - y1
m = -------
   x2 - x1

    - b^2 + 2b - 5 - a^2
m =  --------------------
          b - a

Now we have three expressions for what must be the same slope, so we can write (and solve, we hope) equations:

- b^2 + 2b - 5 - a^2
-------------------- = 2a
       b - a

2a = -2b + 1

Multiply out the first one:

- b^2 + 2b - 5 - a^2 = 2a(b - a)

Simplify the other:
2a = -2b + 1
a = -b + 1 = 1 - b

Now substitute  a = 1 - b into the first:

- b^2 + 2b - 5 - (1 - b)^2 = 2(1 - b)(b - (1 - b))

Remove a few parentheses:

- b^2 + 2b - 5 - (1 - 2b + b^2) = 2(1 - b)(b - 1 + b)
- b^2 + 2b - 5 - 1 + 2b - b^2 = 2(1 - b)(2b - 1)
- b^2 + 2b - 5 - 1 + 2b - b^2 = 2(2b - 1 - 2b^2 + b)
- b^2 + 2b - 5 - 1 + 2b - b^2 = 2(3b - 1 - 2b^2)
- b^2 + 2b - 5 - 1 + 2b - b^2 = 6b - 2 - 4b^2

Now combine terms:

- 2b^2 + 4b - 6 = 6b - 2 - 4b^2
- b^2 + 2b - 3 = 3b - 1 - 2b^2
 b^2 - b - 2 = 0

That's a quadratic, which we can solve, and there will be two solutions:

(b - 2)(b + 1) = 0

b = 2, and so  a = -1
b = -1, and so  a = 2

Now take the first pair,  a = -1,  b = 2 and find the points:

I:  y = x^2.  x = -1, y = 1.

II. y = - x^2 + 2x - 5,  x = 2,  y = -4 + 4 - 5 = -5.

Your line passes through (-1,1) and (2,-5), and its slope is  m = 2a = -2.
Just use your standard "find the equation of a line" stuff to get its equation.

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What about a second line?  Just use the other pair:  a = 2, b = -1.  Go through the same work, get the two points and slope, etc...