Expert: Socrates - 9/27/2006

Question
Hi, I was wondering if you could help me with this one question: Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y=-x^2 +6x -5

Answer
Suppose a line tangent to both curves passes through the points on the curves (a,a^2) and (b,-b^2 + 6b - 5)

The slope of the tangent line to y = x^2 when x=a is 2a

The slope of the tangent line to y = -x^2 + 6x - 5 when x=b is -2b+6

Because these values are both the slope of the same line,

2a = -2b+6

a = -b+3

We know that the common tangent line passes through the points (a,a^2) and (b,-b^2 + 6b - 5).

Using these two points, the slope of the line must be

(-b^2 + 6b - 5 - a^2)/(b-a)


Substitute -b+3 for a in this expression for the slope
and , after simplifying, get

(-2b^2 + 12b - 14) / (2b-3)

But this must be equal to -2b+6 , the slope of the tangent to y = -x^2 + 6x - 5 when x=b .


So now we have

(-2b^2 + 12b - 14) / (2b-3)  = -2b + 6

multiply both sides by 2b-3 and get

-2b^2 + 12b - 14 = (-2b+6)(2b-3)

-2b^2 + 12b - 14 = -4b^2 + 18b - 18

2b^2 - 6b + 4 = 0

b^2 - 3b + 2 = 0

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

so b=1 or b=2

When b=1 ,

a= -b + 3
a = -1 + 3 = 2

Since the line goes through (a,a^2) and (b,-b^2 + 6b - 5)

substitute 2 for a and 1 for b and get

(2,4)  (1,0)

the equation of the line through these points is

y = 4x - 4


When b=2 ,

a= -b + 3
a = -2 + 3 = 1

Since the line goes through (a,a^2) and (b,-b^2 + 6b - 5)

substitute 1 for a and 2 for b and get

(1,1)  (2,3)

the equation of the line through these points is

y = 2x - 1


We now have the answer :

The lines tangent to both curves are

y=4x+4 and y=2x-1