$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{hyperbola} \mathrm{has} \mathrm{one} \mathrm{focus} \mathrm{at} \mathrm{the} \mathrm{origin} \mathrm{and} \mathrm{its} \mathrm{eccentricity} \mathrm{is} \surd 2 \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{directrices} \mathrm{is} \mathrm{x}+\mathrm{y}+1=0 \mathrm{Then} \\ \mathrm{equations} \mathrm{to} \mathrm{its} \mathrm{asymptotes} \mathrm{are} \end{gathered}$$

$$\begin{gathered} By \mathrm{def}inition,equation \mathrm{of} \mathrm{hyperbola} is {\mathrm{x}}^2+{\mathrm{y}}^2=(\surd 2{)}^2((\mathrm{x}+\mathrm{y}+1)/\surd 2{)}^2 \\ \Rightarrow 2\mathrm{xy}+2\mathrm{x}+2\mathrm{y}+1=0 \\ centre of hyperbola=(-1,-1) \\ Now equation of pair of asymptotes is 2\mathrm{xy}+2\mathrm{x}+2\mathrm{y}+k=0 \\ It paeese through the point (-1,-1) then k=2 \\ so required pair of asymptote is 2\mathrm{xy}+2\mathrm{x}+2\mathrm{y}+2=0 \end{gathered}$$