The two parabolas $y^2 = 4a(x+c)$ and $y^2 = 4bx$, where $a>b>0$, cannot have a common normal unless

Explanation

Equation of normal to $y^2=4bx$ is $y=mx-2bm-bm^3$ Equation of normal to $y^2=4a(x+c)$ is $y=m(x+c)-2am-am^3$ For common normal, coefficients must match This gives $2(a-b)m+(a-b)m^3-mc=0$ So $m^2=(c-2(a-b))/(a-b)>0$ Hence $c>2(a-b)$