Common tangent lines of two quadratic functions Find all such lines that are tangent to the following curves:
$$y=x^2$$ and $$y=-x^2+2x-2$$
I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.
 A: Here is a hint for a method which avoids calculus:
The line $y=ax+b$ is a tangent to a quadratic such as $y=x^2$ if and only if the quadratic equation you get by solving these equations simultaneously has a double root. This will give you an equation which must be satisfied by the unknowns $a$ and $b$.
You can do the same for the line $y=ax+b$ and your other quadratic, then solve the two simultaneous equations to find $a$ and $b$.
A: If $y=x^2$ we have that $\frac {dy}{dx} = 2x$
This gives the gradient of the tangent at the point $(x, y)=(a, a^2)$
If the tangent line is $y=mx + c$ we therefore have $a^2=(2a)\cdot a+c$ whence $c=-a^2$ and the general tangent line to $y=x^2$ is $$y=2ax-a^2$$
If $y=-x^2+2x-2$ we have $\frac {dy}{dx} = -2x+2$, so that the tangent line $y=mx +c$ at $(x, y)=(b, -b^2+2b-2)$ is $-b^2+2b-2=(-2b+2)b+c$ whence $c=b^2-2$ and the general tangent to the parabola is $$y=(-2b+2)x+b^2-2$$
If the two tangents are to be the same line we equate coefficients to give:$$a=1-b$$ and $$a^2=2-b^2$$
A: Tangent line of first equation through some point $(x_1,f_1(x_1))$ is
$$
y = f_1(x_1) + f'_1(x_1)(x-x_1) = x_1^2 + 2x_1(x-x_1) = 2x_1x-x_1^2
$$
Tangent line of second equation through some point $(x_2, f_2(x_2))$ is
$$
y = f_2(x_2) + f'_2(x_2)(x-x_2) = -x_2^2+2x_2-2 + (-2x_2+2)(x-x_2)
 = 2(1-x_2)x+x_2^2-2
$$
In order these two lines to be the same one must require
$$
2x_1 = 2(1-x_2) \\
-x_1^2 = x_2^2-2
$$
It is quite simple to solve so I leave it to you. Solution is
$$
k = 1 \pm \sqrt 3 \\
b = -1 \pm \frac {\sqrt 3}2
$$
for the line $y = kx + b$.

A: 
Let $f(x) = x^2$  and  $g(x) = -x^2 + 2x - 2 \implies \dfrac{d}{dx}(f(x))|_{x = a} = 2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = -2b + 2$
$2a = -2b + 2 \implies a = 1 - b \implies$
$A: (1 - b, (1 - b)^2), B: (b, -b^2 + 2b - 2) 
\implies m_{AB} = \dfrac{2b^2 - 4b + 3}{1 - 2b} = -2b + 2 \implies $
$2b^2 - 2b - 1 = 0 \implies b = \dfrac{1 \pm \sqrt{3}}{2}$
Using $b = \dfrac{1 + \sqrt{3}}{2} \implies $
$A: (\dfrac{1 - \sqrt{3}}{2},\dfrac{2 - \sqrt{3}}{2}), B: (\dfrac{1 + \sqrt{3}}{2},\dfrac{\sqrt{3} - 4}{2}) \implies m_{AB} = 1 - \sqrt{3} \implies $
$ \boxed{y = (1 - \sqrt{3})x + \dfrac{\sqrt{3} - 2}{2}} $ 
Using $b = \dfrac{1 - \sqrt{3}}{2} \implies A^{'}: (\dfrac{1 + \sqrt{3}}{2}, \dfrac{2 + \sqrt{3}}{2}), B^{'}: (\dfrac{1 - \sqrt{3}}{2}, \dfrac{-\sqrt{3} - 4}{2}) \implies$
$m_{A^{'}B^{'}} = 1 + \sqrt{3} \implies \boxed{y = (1 + \sqrt{3})x - \dfrac{\sqrt{3} + 2 }{2}}. $ 
