$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$ $\Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$ $\Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {{x \over {\sqrt {1 - {x^2}} }}} \right)} \right) = k$ $\Rightarrow {{\sqrt {1 - 2{x^2}} } \over {\sqrt {1 - {x^2}} }} = k$ $\Rightarrow {{1 - 2{x^2}} \over {1 - {x^2}}} = {k^2}$ $\Rightarrow 1 - 2{x^2} = {k^2} - {k^2}{x^2}$ $\therefore$ ${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)$ ...... (1) and ${\alpha \over \beta } = - 1$ ...... (2) $\therefore$ $2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)( - 1) = - 5$ $\Rightarrow {k^2} = {1 \over 3}$ and $b = S.R = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right) - 1 = 4$ $\therefore$ ${b \over {{k^2}}} = {4 \over {{1 \over 3}}} = 12$