JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Hyperbola

The curve $x^2{y}^2=1$ is symmetric about the origin, so take the square centred there. Its vertices can then be written $(p,q)$, $(-q,p)$, $(-p,-q)$ and $(q,-p)$, and all four lie on the curve exactly when $p^2{q}^2=1$. The midpoint of the side joining $(p,q)$ and $(-q,p)$ is $\left(\frac{p-q}{2},\frac{p+q}{2}\right)$, and it lies on the curve when

Writing $u=p^2$ and $v=q^2$ gives $uv=1$ and $(u-v{)}^2=16$, so

The side joining $(p,q)$ to $(-q,p)$ has squared length $(p+q{)}^2+(p-q{)}^2=2(u+v)=4\sqrt{5}$, and that squared length is the area of the square. Its square is ${\left(4\sqrt{5}\right)}^2=80$.