Is (are) there any straight line(s) that is (are) simultaneously tangent to both the curves given by the equations $y = x^2$ and $y = -x^2+2x-2$?
My Attempt:
Let $y = mx+b$ be any such (non-vertical) line. Then we must have $$ mx+b = x^2 \qquad \mbox{ and } \qquad mx+b = -x^2+2x-2, $$ or in other words, $$ x^2 - mx-b = 0 \qquad \mbox{ and } \qquad x^2 + (m-2)x + (b+2) = 0. $$ Moreover, each one of these two quadratics must have coincident solutions, which implies that each one of the two discriminants must be zero, that is, we must have $$ m^2 + 4b = 0 \qquad \mbox{ and } \qquad (m-2)^2 -4 (b+2) = 0, $$ and from the first of the preceding two relations we get $$ b = - \frac{m^2}{4}, $$ which when substituted into the second one gives $$ (m-2)^2 + m^2 - 8 = 0, $$ which simplifies to $$ 2m^2 -4m -4 = 0, $$ and hence $$ m^2 - 2m - 2 = 0, $$ from which we get $$ m = \frac{ 2 \pm \sqrt{4 + 8} }{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}. $$ For $m = 1 + \sqrt{3}$, we get $$ b = - \frac{ (1 + \sqrt{3})^2 }{4} = - \frac{ 4 + 2 \sqrt{3} }{4} = -1 - \frac{ \sqrt{3} }{2}. $$
And, for $m = 1- \sqrt{3}$, we get $$ b = - \frac{ (1 - \sqrt{3})^2 }{4} = - \frac{ 4 - 2 \sqrt{3} }{4} = -1 + \frac{ \sqrt{3} }{2}. $$
Thus there are two straight lines that are simultaneously tangent to both the given curves, and these lines are given by the equations $$ y = (1 + \sqrt{3} ) x -1 - \frac{ \sqrt{3} }{2} $$ and $$ y = (1 - \sqrt{3} ) x -1 + \frac{ \sqrt{3} }{2}. $$
Is my solution / answer correct? If so, then is my approach to finding this solution / answer also correct? If not, then where have I erred?
Any alternative solutions, especially those using the methods of differential calculus?
To determine the tangent lines is sufficient to solve for $x_1,x_2,y_1,y_2$ the system
$$ \cases{ y_1 = x_1^2\\ y_2 = -x_2^2+2x_2-2\\ y_1 = m x_1 + c\\ y_2 = m x_2 + c } $$
so a typical result is
$$ \left\{ \begin{array}{rcl} x_1& = & \frac{1}{2} \left(m-\sqrt{4 c+m^2}\right) \\ y_1 & = & \frac{1}{2} m \left(m-\sqrt{4 c+m^2}\right)+c \\ x_2 & = & \frac{1}{2} \left(2-\sqrt{(m-4) m-4c-4}-m\right) \\ y_2 & = & c-\frac{1}{2} m \left(\sqrt{(m-4) m-4c-4}+m-2\right) \\ \end{array} \right. $$
from here, we solve the tangency conditions
$$ \cases{ 4 c+m^2 = 0\\ (m-4) m-4c-4=0 } $$
obtaining
$$ \cases{ m = 1\pm \sqrt{3}\\ c=-\frac 14\left(1\pm\sqrt{3}\right)^2 } $$