# Thread: pair of tangents

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

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

$\displaystyle 0 = 1 + a + b$

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

similarly, for $\displaystyle y_2$, the following happens:

$\displaystyle 0 = c - 1$

$\displaystyle \Rightarrow \boxed {c = 1}$

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

and

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

Now we equate the two slopes:

$\displaystyle \Rightarrow 2 + a = c - 2$

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

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

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

$\displaystyle \Rightarrow b = -1 - a$

$\displaystyle \Rightarrow \boxed {b = 2}$

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

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

#### Search Tags

pair, tangents 