Expert: Alon Mandes - 12/31/2008

Question
Please,
1. The tangent point between g(x)=ax^2+bx+c and f(x)=mx+n is x1 resp y1. What's the slope of f(x)=mx+n? (m,n=?)
and vice versa:
2. If I have f(x)= mx+n what is a,b,c=? as long as g(x)=ax^2+bx+c is tangent in x1,y1
Thanks a lot

Answer
Lets find a relation between {x1,y1,m,n,a,b,c} :
1st of all if f(x)=mx+n is a tangent line to the curve g(x) then
m is the slope of the line & it represents the derivative of the
curve in point x1. Meaning : m=g'(x1)=2ax1+b. To find n , we
substitute the point (x1,y1) in f(x) : y1=mx1+n -> n=y1-mx1. So,
1. m=2ax1+b
2. n=y1-ax1^2-bx1
We also know that the point (x1,y1) is on the curve g(x) : so we may
claim that :
3. y1=ax1^2+bx1+c.
Now we have all the relations between the constants. & therefore if some are given then using equations 1,2,3 we can calculate the others.

For example : if m,n & x1,y1 are given then we can calculate a,b,c
as follow:
from equation 1 we can get : b=m-2ax1. substituting this into
equation 2 gives : n=y1-ax1^2-(m-2ax1)x1 -> n=y1-ax1^2-mx1-2ax1^2
-> n=y1-a[x1^2-2x1^2]-mx1 -> n=y1-mx1+ax1^2 -> a=(n-y1+mx1)/x1^2.
Now we need to find b : from equation 1 : b=m-2x1*(n-y1+mx1)/x1^2.
b=m-(2/x1)*(n-y1+mx1).
To find c , we use equation 3 :  y1=ax1^2+bx1+c ->
y1=(x1^2)(n-y1+mx1)/x1^2 +x1[m-(2/x1)*(n-y1+mx1)] + c ->
c=y1-n+y1-mx1+x1m+2(n-y1+mx1)=2y1+2n-2y1+2mx1=2(n+mx1).

Alon.