How to find a tangent line to two different funtions?

Question

Here is the question: Find the equation of a line that is tangent to both of the curves y = x^2 and y = x^2-2x +2 (at different points).

Thanks in advance :) I have no idea where to start lol.

Answer 1

compute the derivatives.

y1' = 2x

y2' = 2x - 2



Through trial and error the the answer is

y = x - 1/4



http://img180.imageshack.us/my.php?image=asfez0.png

Answer 2

let (α, β) be a point on y = x ² , such that the tangent at (α, β) to

y = x ² is also tangent to y = x ² -2x +2



therefore β = α ² -----------(1)



y = x ² implies y' = 2x implies slope of the tangent at (α, β) is given by m = 2α



equation of the required tangent is of the form

( y - β) = 2α(x - α)

or

y = 2α x -2α² +β or



y = 2α x -2α² + α ² using equation(1)



y = 2α x - α ² --------------(2)



now solve (2) with y = x ² - 2x +2 and use the condition for tangency.



solving

2α x - α ² = x ² - 2x +2



or

x ² - 2 (1+α) x + ( 2 + α ² ) = 0



condition for tangency is that the above equation has equal roots, hence discriminant =0

4(1+α) ² - 4(2 + α ²) =0



simplify to get α = ½



use equation (2) to get the equation of the tangent as



y = x - (1/4)

Answer 3

let the first curve is y1 and second curve is y2



let the point of tangency of first curve is x1,y1



and point of tangency of second curve is x2,y2



The slope of tangent is derivative of the curve



find out derivative of the first curve (y1 = x^2)



y1' = 2x



at x = x1 , the slope is 2x1



the derivative of second curve y2= x^2 - 2x + 2



y2' = 2x - 2



at x = x2



y2' = 2x2 - 2



since tangent is same, slope must be equal



so 2x1 = 2x2 - 2



x2 - x1 = 1



y cordinate point of tangency of first curve at x= x1 is



y1 = x1^2



y coordinate point of tangency of second curve at x = x2 is



y2 = x2^2 - 2x2 + 2



so y2 - y1 = x2^2 - 2x2 + 2 - x1^2



y2 - y1 = (x2^2 - x1^2) - 2x2 + 2



=>(x2 - x1)(x2+x1) - 2x2 + 2



=>(x2+x1) - 2x2 + 2 (since x2 - x1 = 1)



=>x1 - x2 + 2 = -(x2 - x1) + 2 = -1 + 2 = 1



so (y2 - y1)/(x2-x1) = 1/1 = 1



so slope is 1



so 2x1 = 1 : x1 = 1/2



2x2 - 2 = 1



2x2 = 3



x2 = 3/2



y1 = (1/2)^2 = 1/4



y2 = (3/2)^2 - 2(3/2) + 2------> 9/4 - 3 + 2



y2 = 5/4



so x1,y1 = (1/2, 1/4) and x2,y2 = (3/2, 5/4)



The equation of tangent is



y - y1 = m(x-x1)



y - 1/4 = 1(x - 1/2)



y = x - 1/2 + 1/4



y = x - 1/4 or 4y = 4x - 1

Answer 4

Okay, so, you need two points to define a line, and given the conditions of the problem, they need to be points with particular properties. Let us call one of them (x1, y1) and the second (x2, y2). They will be the points at which the line is tangent to the curves. To be tangent to a curve, the line has to have a point that's also on the curve, and at that point, the slope of the curve has to be equal to the slope of the line.



So, we know the line containing (x1, y1) and (x2, y2) has slope:

m=(y2-y1)/(x2-x1).



We also know that if (x1,y1) is on y=x^2, then

y1=(x1)^2



and (taking the derivative)



m=2(x1)



--since that's the slope of the curve y=x^2 at (x1,y1), and for tangency we have to find a point where that's equal to the slope of the line.



Similarly, we know that

y2=(x2)^2-2(x2)+2



and



m=2(x2)-2



Now, technically you've got a system of five variables in five equations:



m=(y2-y1)/(x2-x1) -- by definition of the slope of a line

y2=(x2)^2-2(x2)+2 -- to intersect with y=x^2-2x+2

m=2(x2)-2 -- to match slope of y=x^2-2x+2

y1=(x1)^2 -- to intersect with y=x^2

m=2(x1) -- to match slope of y=x^2



From here, you can start making some substitutions. For instance:



m=(y2-y1)/(x2-x1)

--subst. for y2

m=((x2)^2-2(x2)+2-y1)/(x2-x1)

--subst. for y1

m=((x2)^2-2(x2)+2-(x1)^2)/(x2-x1)



Now note that m=2(x1)=2(x2)-2

x1=(x2)-1

x2=(x1)+1

Since there are more terms with x2 already, it's probably simplest to get everything in those terms. So:

2(x2)-2=((x2)^2-2(x2)+2-((x2)-1)^2)/(x2-(x2-1))

2(x2)-2=((x2)^2-2(x2)+2-((x2)^2-2(x2)+1))/(x2-x2+1)

2(x2)-2=1 (seriously, watch the terms cancel out)

2(x2)=3

x2=1.5



Check my work there--the conceptual part is right but I could have messed up in the parentheses someplace--and then you can go through and find the rest of the values using that value. Once you've got the two points, you can find the equation of the line.

Answer 5

set the derivatives equal and solve...?