A square is formed such that one pair of its opposite sides are tangential to the hyperbola $x^2-y^2=1.$ Which of the following is/are true?

Let tangent of solve of slope m be $y=mx\pm \sqrt{m^2-1}$ . Then side length will  be $2\sqrt{\frac{m^2-1}{m^2+1}}$ Also. Length of diagonal of square should be greater than or equal to  distance between points of tangency
 $\Rightarrow 2\sqrt{2}\sqrt{\frac{m_2-1}{m_2+1}}\Rightarrow m^2\ge {(\sqrt{2}+1)}^2$
Side length  $=2\sqrt{\frac{m^2-1}{m^2+1}}$ (for smallest square)  $=2\sqrt{\frac{2+2\sqrt{2}}{4+2\sqrt{2}}}=2^{\frac{3}{4}}$
Side length of largest possible square is  $2\left(\text{as m}\to \infty \right)$