Let $\mathrm{A}(\mathrm{sec\theta }, \sqrt{2}\mathrm{tan\theta })$ and $\mathrm{B}(\mathrm{sec\varphi }, \sqrt{2}\mathrm{tan\varphi })$ where $\mathrm{\theta }+\mathrm{\varphi }=\frac{\mathrm{\pi }}{2},$be two points on the hyperbola $2{\mathrm{x}}^2 – {\mathrm{y}}^2 = 2.$ If $\left(\mathrm{\alpha },\mathrm{\beta }\right)$ is the point of intersection of the normals to the hyperbola at A and B,then ${\left(2\mathrm{\beta }\right)}^2$ is equal to $\mathrm{\_\_\_\_\_}$

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{1}-\frac{{\mathrm{y}}^2}{2}=1 \\ \Rightarrow {\mathrm{a}}^2=1, {\mathrm{b}}^2=2 \\ \mathrm{Eeuation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{A}(\mathrm{sec\theta }, \sqrt{2}\mathrm{tan\theta }) \mathrm{is} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{sec\theta }}+\frac{\sqrt{2}\mathrm{y}}{\mathrm{tan\theta }}=3 \\ \Rightarrow \mathrm{x} \mathrm{cos\theta }+\sqrt{2}\mathrm{y} \mathrm{cot\theta }=3 — ① \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{at} (\sec \cancel{\mathrm{\theta }}, \sqrt{2} \tan \cancel{\mathrm{\theta }}) \mathrm{is} \\ \frac{\mathrm{x}}{\sec \cancel{\mathrm{\theta }}}+\frac{\sqrt{ }\mathrm{y}}{\tan \cancel{\mathrm{\theta }}}=3 \Rightarrow \mathrm{x} \cos \cancel{\mathrm{\theta }}+\sqrt{2}\mathrm{y} \cot \cancel{\mathrm{\theta }}=3 \\ \Rightarrow \mathrm{x} \mathrm{sin\theta }+\sqrt{2}\mathrm{y} \mathrm{tan\theta }=3 — ② \left(\because \varnothing =\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right) \\ ① \times \mathrm{sin\theta }-② \times \mathrm{cos\theta } \Rightarrow \sqrt{2} \mathrm{y}(\mathrm{cos\theta }-\mathrm{sin\theta })=3(\mathrm{sin\theta }-\mathrm{cos\theta }) \\ \Rightarrow \mathrm{y}=\frac{-3}{\sqrt{2}} \\ \mathrm{It} (\mathrm{\alpha }, \mathrm{\beta }) \mathrm{is} \mathrm{P}.\mathrm{I} \mathrm{of} ① \& ② \mathrm{then} \mathrm{\beta }=\frac{-3}{\sqrt{2}} \\ \mathrm{Now} {\left(2\mathrm{\beta }\right)}^2=4\times \frac{9}{2}=18 \end{gathered}$$