# Tangent of two curves

1. May 25, 2010

### songoku

1. The problem statement, all variables and given/known data
let C1 : y = x - 1/2 x2 and C2 : x = y - 1/2 y2 be curves on the xy plane.

1. find the equation of the tangent to the curve C1 at x = k

2. suppose the line obtained in 1) is also tangent to the curve C2. find all values of k and the equations of the tangents.

3. evaluate the area of the figure enclosed by all tangents obtained in 2) and the curve C2

2. Relevant equations
derivatives, equation of tangent

3. The attempt at a solution
1. I've done it. I got : y = (1-k) x + 1/2 k2

2.
differentiate C2 with respect to x :
1 = dy/dx - y dy/dx

dy/dx = 1/(1-y) = m

so, 1/(1-y) = 1-k

then.....I...gave up...

2. May 25, 2010

### rock.freak667

If x=k is a point on C1, then the point (k,k-1/2k2) ((x,y)) is a point on the line right?

So the gradient of the tangent on C2 is dy/dx= 1/(1-y). Since (k,k-1/2k2) lies on the line, what is the gradient of this tangent?

then put that equal to 1-k.

3. May 25, 2010

### songoku

Do you mean substituting y = k - 1/2 k2 to dy/dx= 1/(1-y) ?

4. May 25, 2010

### annoymage

sorry (editted)

alternative way,

you also can find another tangent equation with gradient 1/(1-y) with point (k,k-1/2k2), and compare with the other tangent eqution

5. May 25, 2010

### songoku

I don't get it. x = k is the common point between the tangent and C1 and I think we can't use it to find the equation of tangent of C2 since x = k may not be the common point.

Or maybe I missed the hint?

6. May 26, 2010

### annoymage

it says that, the tangent of C1 at x=k , is also tangent C2 at some point x, implies that the tangent of C1 and tangent of C2 is the same line.

since it is the same line, tangent C2 also pass through point (k,k-1/2k2)

Last edited: May 26, 2010
7. May 26, 2010

### annoymage

i'm sorry but i think i'm wrong

Last edited: May 26, 2010