Solve this following

Normal to these two curves are

$y=m(x-c)-2 b m-b m^{3}$

$y=m x-4 a m-2 a m^{3}$

If they have a common normal

$(c+2 b) m+b m^{3}=4 a m+2 a m^{3}$

Now $(4 a-c-2 b) m=(b-2 a) m^{3}$

We get all options are correct for $m=0$ (common normal x-axis)

Ans. $(1),(2),(3),(4)$

If we consider question as

If the parabolas $y^{2}=4 b(x-c)$ and $y^{2}=8 a x$ have a common normal other than $x$-axis, then which one of the following is a valid choice for the ordered triad $(a, b, c)$ ?

When $m \neq 0:(4 a-c-2 b)=(b-2 a) m^{2}$

$m^{2}=\frac{c}{2 a-b}-2>0 \Rightarrow \frac{c}{2 a-b}>2$

Now according to options, option 4 is correct