If the parabolas y ^ 2 = 4b (x - c) and y ^ 2 = 8ax have a common...

Axis of both parabola is $x$-axis i.e. $y = 0$.

As we know, axis of parabola is always a normal to the parabola.

Hence, for all values of $\left(a, b, c\right)$ a common normal is present.

Another Method:
Normal to parabola ${y}^{2} = 4ax$ is give by

$y = mx - 2am - a{m}^{3}$

$∴$Normal to parabola

${y}^{2} = 4b\left(x - c\right) is</p><p>$y = m\left(x - c\right)$- 2bm - b{m}^{3}$

\left[replacingabybandxbyx - c\right]

$= mx - \left(2b + c\right)m - b{m}^{3} . . . \left(1\right)</p><p>and normal$ to parabola{y}^{2} = 8ax$is$. . . \left(2\right)$</p><p>$[ replacingaby2a ]$</p><p>For common normal, we should have$mx - 4am - 2a{m}^{3} = mx - \left(2b + c\right)m - b{m}^{3}$</p><p>[using Equations. \left(1\right)$ and \left(2\right)$]</p><p>$4am + 2a{m}^{3} = \left(2b + c\right)m + b{m}^{3}$</p><p>$⇒ m\left(\left(2a - b\right){m}^{2} + \left(4a - 2b - c\right)\right) = 0$</p><p>$⇒ m = 0$</p><p>or${m}^{2} = \dfrac{2b + c - 4a}{2a - b}$</p><p>$= \dfrac{c}{2a - b} - 2$</p><p>As,${m}^{2} > 0,$therefore$\dfrac{c}{2a - b} > 2$</p><p>Note that if$m = 0,$then all options satisfy (:$y = 0 is a common normal).