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

Answer: 38

$\frac{t_1+t_2}{2}\cdot \frac{\frac{1}{t_1}-\frac{1}{t_2}}{2}=1$
$\Rightarrow t_1^2-t_2^2=4t_1{t}_2$
$\frac{1}{t_1^2}\times \left(-\frac{1}{t_2^2}\right)=-1\Rightarrow t_1{t}_2=1$
$\Rightarrow {\left(t_1{t}_2\right)}^2=1\Rightarrow t_1{t}_2=1$
$t_1^2-t_2^2=4$
$\Rightarrow t_1^2+t_2^2=\sqrt{4^2+4}=2\sqrt{5}$
$\Rightarrow t_1^2=2+\sqrt{5}\Rightarrow \frac{1}{t_1^2}=\sqrt{5}-2$
$A{B}^2={\left(t_1-t_2\right)}^2+{\left(\frac{1}{t_1}+\frac{1}{t_2}\right)}^2$
$=2\left(t_1^2+\frac{1}{t_1^2}\right)=4\sqrt{5}\Rightarrow Area{a}^2=80$