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

$\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=4 t_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 $
$A B^2=\left(t_1-t_2\right)^2+\left(\frac{1}{t_1}+\frac{1}{t_2}\right)^2$
$=2\left(t_1^2+\frac{1}{t_1^2}\right)=4 \sqrt{5} \Rightarrow A r e a a^2=80$