1. ## pair of tangents

Here's the question:

I don't know how to proceed.

2. Originally Posted by cinder
Here's the question:

I don't know how to proceed.
First thing we do know about these curves, is that both have the point (1,0)

Let $y_1 = x^2 + ax + b$
Let $y_2 = cx - x^2$

We have, when x = 1, y = 0. so for $y_1$ the following happens:

$0 = 1 + a + b$

$\Rightarrow a + b = -1$ ..............(1)

similarly, for $y_2$, the following happens:

$0 = c - 1$

$\Rightarrow \boxed {c = 1}$

So we know the value of c, and we know one equation that relates a and b. but what else do we know?

they share a common tangent at (1,0), in particular, the slope of the tangent line at x = 1 is the same for the two curves

so, $y_1 ' = 2x + a$, when x = 1, we get: $y_1 ' = 2 + a$

and

$y_2 ' = c - 2x$, when x = 1, we get: $y_2 ' = c - 2$

Now we equate the two slopes:

$\Rightarrow 2 + a = c - 2$

$\Rightarrow a = c - 4$ ...............(2)

$\Rightarrow \boxed {a = -3}$ ............since $c = 1$

But from equation (1), $a + b = -1$

$\Rightarrow b = -1 - a$

$\Rightarrow \boxed {b = 2}$

So, $\boxed {y_1 = x^2 -3x + 2} \mbox { and } \boxed {y_2 = x - x^2}$

3. For $a$ and $b$ we can do in this way:
The two curves have one point of intersection (the tangent point) so the equation $x^2+ax+b=cx-x^2$ must have one solution (or equal roots). The equation can be written as $2x^2+(a-c)x+b=0$.
Then $\triangle =(a-c)^2-8b=(a-1)^2-8b=0$.
Solving the system $\left\{\begin{array}{l}a+b=-1\\(a-1)^2-8b=0\end{array}\right.$ we have $a=-3,b=2$