A square A B C D has all its vertices on the curve x 2 y 2 = 1 . The midpoints of its sides also lie on the…

$xy=1, -1$

$\frac{t_1+t_2}{2}·\frac{\frac{1}{t_1}-\frac{1}{t_2}}{2}=1$

$\Rightarrow t_2^2-t_1{}^2=4t_1{t}_2$

Product of slope $=-1$

$\frac{1}{t_1^2}\times \left(-\frac{1}{t_2^2}\right)=-1\Rightarrow t_1{t}_2=1$

$\Rightarrow {\left(t_1{t}_2\right)}^2=1\Rightarrow t_1{t}_2=1$

$\Rightarrow t_1^2+t_2^2=\sqrt{4^2+4}=2\sqrt{5}$

$\Rightarrow t_1^2=2+\sqrt{5}\Rightarrow \frac{1}{t_1^2}=\sqrt{5}-2$

$A{B}^2={\left(t_1-t_2\right)}^2+{\left(\frac{1}{t_1}+\frac{1}{t_2}\right)}^2$

$=2\left(t_1^2+\frac{1}{t_1^2}\right)=4\sqrt{5}\Rightarrow {\text{ Area }}^2=80$ sq. unit