Question

A square $ABCD$ has all its vertices on the curve $x^2{y}^2=1.$ The midpoints of its sides also lie on the same curve. Then, the square of area of $ABCD$ is

Answer 1

OA $\perp$ OB
$\Rightarrow \left(\frac{1}{p^2}\right)+(-\frac{1}{q^2})=1$
$\Rightarrow p^2{q}^2=1$

$p(\frac{p+q}{2},\frac{\frac{1}{p}-\frac{1}{q}}{2})$ lies on $x^2{y}^2=1$
$\Rightarrow (p+q{)}^2{(\frac{1}{p}-\frac{1}{q})}^2=16$
$\Rightarrow (p+q{)}^2(p-q{)}^2=16$
$\Rightarrow (p^2-q^2{)}^2=16$
$\Rightarrow p^2-\frac{1}{p^2}=\pm 4$
$\Rightarrow p^2\pm 4p^2-1=0$
$\Rightarrow p^2=\frac{\pm 4\pm \sqrt{20}}{2}=\pm 2\pm \sqrt{5}$
$\Rightarrow p^2=2+\sqrt{5}$ or $-2+\sqrt{5}$

$O{B}^2=p^2+\frac{1}{p^2}=2+\sqrt{5}+\frac{1}{2+\sqrt{5}}\text{or}-2+\sqrt{5}+\frac{1}{-2+\sqrt{5}}=2\sqrt{5}$
Area $=4\left(\frac{1}{2}\right)\left(OA\right)\left(OB\right)=2(OB{)}^2=4\sqrt{5}$