Expert: Paul Klarreich - 10/29/2005

Question
find the equations of the two tangent lines that are tangent to both lines y=x^2 and y= -x^2+6x-5.

they need two lines. each line is tangent to a point on each graph. i got the answer using guess and check but my teacher says i should find the calculus way to find the lines and the points they are tangent to.

this isn't for any extra points just as an exercise which i can't figure out so please feel free to completely show me how this is done because it is killing me not knowing. thank you

Answer
Hi, Matt,

Find the equations of the two lines that are tangent to both graphs

(A) y = f(x) = x^2 and
(B) y = g(x) = -x^2+6x-5.

You need two lines. each line is tangent to a point on each graph. i got the answer using guess and
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Sorry I don't have a brilliant and elegant solution for you.  The following klutzy solution will have to do.  

Suppose  we have such a line, which we call C.  Then there is a point (x1,y1) on graph A and one (x2,y2) on graph B, that satisfy these conditions:

(Sorry I don't have an easy way to write subscripts.  You will have to understand that 'x1' means 'x-sub-1'.

1. (x1,y1) is on both A and C.
2. (x2,y2) is on both B and C.
3. The slope of the tangent, a, is the same as f'(x1) = 2x1
4. The slope of the tangent, a, is the same as g'(x2) = -2x2 + 6

We just use the point-slope form of the equation of a line to write the two equations, which ARE EQUATIONS OF THE SAME LINE and do some algebra.  Don't worry about the fact that you know there are two solutions -- that will take care of itself.

For the first:  m = 2x1, and y1 = x1^2; the equation is:

y - y1 = m(x - x1)

y - x1^2 = 2x1(x - x1) = 2xx1 - 2x1^2

y = 2xx1 - x1^2   <====  First form.

For the second: m = -2x2 + 6, y2 = -x2^2 + 6x2 - 5

y - (-x2^2 + 6x2 - 5) = (-2x2 + 6)(x - x2)

y + x2^2 - 6x2 + 5 = -2xx2 + 2x2^2 +6x - 6x2

y  = -2xx2 + x2^2 +6x - 5

y = (-2x2 + 6)x + x2^2 - 5  <====== Second form.

Since these are the same equation, the coefficients of x and the constant terms must be the same:

2x1 = -2x2 + 6   AND  -x1^2 = x2^2 - 5  or x1^2 = -x2^2 + 5
So we do a little algebra to solve these two simultaneous equations for x1 and x2:

x1 = -x2 + 3;  put that into the second eqn:
(-x2 + 3)^2 = -x2^2 + 5
x2^2 -6x2 + 9 = -x2^2 + 5

2x2^2 - 6x2 + 4 = 0
x2^2 - 3x2 + 2 = 0
Factor and solve:

(x2 - 2)(x2 - 1) = 0,
So  x2 = 2 and x2 = 1 are the solutions.

The rest you should be able to handle:
1. Substitute these to get two values of x1.
2. Take each pair (x,y) and write the equation of the line satisfying the m,x,y conditions.