The equation of a common tangent to the parabolas $\mathrm{y}={\mathrm{x}}^2\text{ and }\mathrm{y}=-(\mathrm{x}-2{)}^2$ is

Equation of tangent of $\mathrm{y}={\mathrm{x}}^2$ be

$\begin{array}{l}\mathrm{tx}=\mathrm{y}+{\mathrm{at}}^2 ...\left(1\right)\\ \mathrm{y}=\mathrm{tx}-\frac{{\mathrm{t}}^2}{4}\end{array}$

Solve with $\mathrm{y}=-(\mathrm{x}-2{)}^2$

$\begin{array}{l}\mathrm{tx}-\frac{{\mathrm{t}}^2}{4}=-(\mathrm{x}-2{)}^2\\ {\mathrm{x}}^2+\mathrm{x}(\mathrm{t}-4)-\frac{{\mathrm{t}}^2}{4}+4=0\\ \mathrm{D}=0\\ (\mathrm{t}-4{)}^2-4\cdot \left(4-\frac{{\mathrm{t}}^2}{4}\right)=0\\ {\mathrm{t}}^2-4\mathrm{t}=0\\ \mathrm{t}=0\text{ or }\mathrm{t}=4\end{array}$

From eq. (1), required common tangent is

y = 4 (x - 1)