A square ABCD has all its vertices on the cur | vedclass

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(D) The curve is $x^{2}y^{2}=1$,which implies $xy=1$ or $xy=-1$. Let the vertices be $A(t_1, 1/t_1)$,$B(t_2, -1/t_2)$,$C(-t_1, -1/t_1)$,and $D(-t_2, 1

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$80$

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(D) The curve is $x^{2}y^{2}=1$,which implies $xy=1$ or $xy=-1$. Let the vertices be $A(t_1, 1/t_1)$,$B(t_2, -1/t_2)$,$C(-t_1, -1/t_1)$,and $D(-t_2, 1/t_2)$.Since $ABCD$ is a square,the midpoint of $AB$ must lie on the curve $xy=1$ or $xy=-1$. The midpoint of $AB$ is $M = (\frac{t_1+t_2}{2}, \frac{1/t_1 - 1/t_2}{2})$.For $M$ to lie on $xy=1$,we have $(\frac{t_1+t_2}{2})(\frac{t_2-t_1}{2t_1t_2}) = 1$,which simplifies to $t_2^2 - t_1^2 = 4t_1t_2$.Also,the slope of $AB$ must be $-1$ (since $ABCD$ is a square and the vertices are symmetric about the origin). The slope of $AB$ is $\frac{-1/t_2 - 1/t_1}{t_2 - t_1} = \frac{-(t_1+t_2)}{t_1t_2(t_2-t_1)} = -1$,which implies $t_1+t_2 = t_1t_2(t_2-t_1)$.Solving these equations,we find $t_1t_2 = 1$ and $t_2^2 - t_1^2 = 4$. Thus $t_2^2 + t_1^2 = \sqrt{4^2 + 4(1)^2} = \sqrt{20} = 2\sqrt{5}$.The side length squared is $AB^2 = (t_2-t_1)^2 + (-1/t_2 - 1/t_1)^2 = (t_2-t_1)^2 + (\frac{t_1+t_2}{t_1t_2})^2 = (t_2^2+t_1^2 - 2t_1t_2) + (t_1+t_2)^2 = (2\sqrt{5}-2) + (2\sqrt{5}+2) = 4\sqrt{5}$.Area of $ABCD = AB^2 = 4\sqrt{5}$.Therefore,the square of the area is $(4\sqrt{5})^2 = 16 	imes 5 = 80$.

$A$ 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 the area of $ABCD$ is

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