The equation of a common tangent to the parabolas y = x ^ 2 and y =...

Let $y = mx + c$ be the tangent to $y = {x}^{2}$, then

${x}^{2} - mx - c = 0$

For tangency, discriminant $D = 0$

$⇒ {m}^{2} = - 4c \ldots \left(1\right)</p><p>Now let$y = mx + c$be also tangent to$y = - {$\left(x - 2\right)$}^{2}$, then</p><p>${\left(x - 2\right)}^{2} + mx + c = 0

For $D = 0$, we get

${m}^{2} - 8m = 4c \ldots \left(2\right)$

From \left(1\right)$and \left(2\right)$, we get

$m = 0, c = 0 or m = 4, c = - 4$

So, there will be two common tangents

y = 0 or y = 4x - 4