Tangent line at point P

Question

so I have this question here.

''We say two curves are tangent to each other at a point $P$, if both of these curves pass through $P$ and have the same tangent line at $P$. Find $a$, $b$ and $c$ (if any) such that the curves $y=x^2+ax+b$ and $y=cx-x^2$ are tangent to each other at the point $(1,0)$.''

First of all, what is this even asking? The curves should be tangent to each other right? Does that mean both curves should touch each other a exactly one point?

Second, how would I go about doing this? I have a good idea but I am not sure if it's right. The tangent line slopes should be equal to each other so I was going to take the derivative and set them equal to each other.

The problem is that I don't know if $(1,0)$ are common to both curves so I can't plug those values in to get some sort of system of equations.

Any guidance on this?

Answer 1

Hint:

The problem say that $(1,0)$ must be a common point of the two curves, because they have here a common tangent. So we have:

From the first: $ 1+a+b=0 \Rightarrow b=-1-a $

from the second $ c-1=0 \Rightarrow c=1 $

and the two parabolas becomes: $$ y=x^2+ax-a-1 \qquad y=x-x^2 $$

can you do from this?

The figure illustrate the situation.

Answer 2

Indeed the functions themselves must satisfy the point (1,0).

Two functions $f(x)$ and $g(x)$ are said to be tangent to each other at some point $(a,b)$ iff $f(a)= g(a) = b$ and $f'(a) = g'(a)$. Using these 2 conditions, you can easily solve linear equations in a,b,c for your problem.