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Step-by-Step Solution
Step 1: Understand the curve and the square
We are given the curve
x^2 y^2 = 1 .
Rewriting,
y^2 = \frac{1}{x^2}
so
y = \pm \frac{1}{x} .
This represents a rectangular hyperbola (specifically, two branches in the plane).
Step 2: Represent the vertices on the curve
Let the square be ABCD . From the provided construction in the reference solution, the vertices lie on the hyperbola, and the midpoints of each side also lie on the same hyperbola.
Step 3: Key property used (orthogonality of adjacent sides)
For ABCD to be a square, adjacent sides must be perpendicular. If we set up coordinates of two adjacent vertices (from the origin-based approach in the solution) as (p, \tfrac{1}{p}) and (q, -\tfrac{1}{q}) , their respective slopes multiply to -1 for perpendicularity. This leads to the condition
p^2 q^2 = 1 .
Step 4: Midpoints lying on the hyperbola
The midpoint P of side AB , using coordinates (p,\tfrac{1}{p}) and (q,-\tfrac{1}{q}) , is
\displaystyle \left(\frac{p+q}{2}, \frac{\tfrac{1}{p} - \tfrac{1}{q}}{2}\right).
Since P also lies on x^2y^2=1, it imposes:
\[
\biggl(\frac{p+q}{2}\biggr)^2 \,\biggl(\frac{1/p - 1/q}{2}\biggr)^2 =1.
\]
This simplifies to a relationship involving p and q that yields
(p+q)^2 (p-q)^2 =16.
Step 5: Simplify and solve for p^2
Further algebraic steps reduce to
\[
(p^2 - q^2)^2 =16
\quad \text{and} \quad
p^2q^2 =1.
\]
By substituting q^2 = \frac{1}{p^2} and rearranging, one obtains a quadratic in p^2 :
\[
p^4 \pm 4\,p^2 -1 =0
\quad \Longrightarrow \quad
p^2 =2 + \sqrt{5}
\quad \text{or} \quad
p^2 =-2+\sqrt{5}.
\]
Usually, we take the positive viable solution p^2 =2+\sqrt{5}.
Step 6: Find the side length and the area of the square
From p^2 =2+\sqrt{5}, one evaluates
\[
OB^2 =p^2 +\frac{1}{p^2} =2+\sqrt{5} +\frac{1}{2+\sqrt{5}}.
\]
After rationalizing, this simplifies to OB^2 =2\sqrt{5}. Hence
OB =\sqrt{2\sqrt{5}} = \sqrt{2}\,\sqrt[4]{5}.
In the reference construction, the square’s side can be deduced from such segments, and ultimately the area is determined to be
\[
\text{Area} =4\sqrt{5}.
\]
Step 7: Final answer – square of the area
The problem specifically asks for the square of the area of ABCD . Since the area is
4\sqrt{5},
its square is
\[
(4\sqrt{5})^2 = 16 \times 5 =80.
\]
Therefore, the square of the area of the square ABCD is 80.
