If the normals drawn to the hyperbola xy=4 at $( {\mathrm{\alpha }}_{\mathrm{i}},{\mathrm{\beta }}_{\mathrm{i}} ) (\mathrm{i} = 1, 2, 3, 4)$ are concurrent at the point (a, b) then $\frac{\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_3+{\mathrm{\alpha }}_4\right)}{\left({\mathrm{\beta }}_1+{\mathrm{\beta }}_2+{\mathrm{\beta }}_3+{\mathrm{\beta }}_4\right)}\left({\mathrm{\alpha }}_1{\mathrm{\alpha }}_2{\mathrm{\alpha }}_3{\mathrm{\alpha }}_4\right)=$
 

Given hyperbola xy=4

let ($2t,\frac{2}{t}) be any point on given curve$

Now equation of normal at 't' is $\mathrm{y}-\frac{\mathrm{c}}{\mathrm{t}}={\mathrm{t}}^2(\mathrm{x}-\mathrm{ct})$

If it passes through (a, b) then

$$\begin{gathered} \Rightarrow \mathrm{b}-\frac{\mathrm{c}}{\mathrm{t}}={\mathrm{t}}^2(\mathrm{a}-\mathrm{ct}) \\ \Rightarrow \mathrm{bt}-\mathrm{c}={\mathrm{t}}^3(\mathrm{a}-\mathrm{ct}) \\ \Rightarrow \mathrm{bt}-\mathrm{c}={\mathrm{at}}^3-{\mathrm{ct}}^4 \\ \Rightarrow {\mathrm{ct}}^4-{\mathrm{at}}^3+\mathrm{bt}-\mathrm{c}=0 \\ \Rightarrow 2{\mathrm{t}}^4-{\mathrm{at}}^3+\mathrm{bt}-2=0 (\because \mathrm{where} {\mathrm{c}}^2=4 \Rightarrow \mathrm{c}=2) \\ \mathrm{let} {\mathrm{t}}_1, {\mathrm{t}}_2, {\mathrm{t}}_3, {\mathrm{t}}_4 be \mathrm{four} \mathrm{roots}. \\ {\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3+{\mathrm{t}}_4=\frac{\mathrm{a}}{2}, \sum _{ }^{ }{t}_1{\mathrm{t}}_2=\mathrm{O}, \sum _{ }^{ }{t}_1{\mathrm{t}}_2{\mathrm{t}}_3=\frac{-\mathrm{b}}{2}, \mathrm{and} {\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3{\mathrm{t}}_4=-1. \end{gathered}$$$$\begin{gathered} \mathrm{Now} {\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_3+{\mathrm{\alpha }}_4=(2{\mathrm{t}}_1+2{\mathrm{t}}_2+2{\mathrm{t}}_3+2{\mathrm{t}}_4) \\ =2\left(\frac{\mathrm{a}}{2}\right)=\mathrm{a} \\ {\mathrm{\alpha }}_1.{\mathrm{\alpha }}_2.{\mathrm{\alpha }}_3.{\mathrm{\alpha }}_4=\left(2{\mathrm{t}}_1\right) \left(2{\mathrm{t}}_2\right) \left(2{\mathrm{t}}_3\right) \left(2{\mathrm{t}}_4\right)=16(-1)=-16 \\ \mathrm{Now} {\mathrm{\beta }}_1+{\mathrm{\beta }}_2+{\mathrm{\beta }}_3+{\mathrm{\beta }}_4=\frac{2}{{\mathrm{t}}_1}+\frac{2}{{\mathrm{t}}_2}+\frac{2}{{\mathrm{t}}_3}+\frac{2}{{\mathrm{t}}_4} \\ =2\left(\frac{1}{{\mathrm{t}}_1}+\frac{1}{{\mathrm{t}}_2}+\frac{1}{{\mathrm{t}}_3}+\frac{1}{{\mathrm{t}}_4}\right)=2\left[\frac{{\displaystyle \sum _{ }^{ }}{t}_1{\mathrm{t}}_2{\mathrm{t}}_3}{{\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3{\mathrm{t}}_4}\right] \\ =\frac{2\left({\displaystyle \frac{-\mathrm{b}}{2}}\right)}{-1}=\mathrm{b} \\ \therefore \mathrm{Req}. \mathrm{value} =\frac{-16\mathrm{a}}{\mathrm{b}} \end{gathered}$$