Question

Sol. [Key Concept: A normal $y = - tx + 2at + {at}^{3}$ common to both parabolas forces the same slope-parameter $t$; consistency gives the condition on $|\frac{2(a - a')}{\beta' - \beta}|$.]

Eqn. of $AB$ is

$$\begin{array}{r} y - \alpha = - t(x - \beta) + 2at + at^{3}\#(1) \end{array}$$

or $y - \alpha = - t^{'}\left( x - \beta^{'} \right) + 2a^{'}t^{'} + a^{'}t^{'3}$

(1) and (2) are identical

$$\Rightarrow \ t = t^{'}\text{~}\text{and}\text{~}2at + at^{3} + \beta t + \alpha$$

$$= 2a^{'}t + a^{'}t^{3} + t\beta^{'} + \alpha$$

$$\begin{matrix} & \\ & \\ \Rightarrow \ t^{2} = \frac{\left( \beta^{'} - \beta \right) + 2\left( a^{'} - a \right)}{\left( a - a^{'} \right)} & \text{(3)} \\ \Rightarrow \ \frac{\beta^{'} - \beta}{a - a^{'}} > 2 \Rightarrow \frac{2\left( a - a^{'} \right)}{\left( \beta^{'} - \beta \right)} < 1 & \text{(3)} \end{matrix}$$