Question

If the parabolas $y^2=4b(x-c)$ and $y^2=8ax$ have a common normal, then which one of the follwing is a valid choice for the ordered triplets $(a,b,c)$?

Answer 1

If $m$ is the slope of the common normal, then the equation of normal to parabola $y^2=4b(x-c)$ is given by
$y=m(x-c)-2bm-b{m}^3\cdots \left(1\right)$

Equation of normal to parabola $y^2=8ax$ is
$y=mx-4am-2a{m}^3\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$m(x-c)-2bm-b{m}^3=mx-4am-2a{m}^3$
$\Rightarrow m^2(2a-b)=c-2(2a-b)$
$\Rightarrow m^2=\frac{c}{2a-b}-2\because m^2\ge 0\Rightarrow \frac{c}{2a-b}-2\ge 0$
$\therefore \frac{c}{2a-b}\ge 2\cdots \left(3\right)$
Hence $(1,1,3)$ satisfies equation $\left(3\right)$