Solutions:

\( OA \perp OB \)
\( -\frac{1}{a^2} \cdot \frac{1}{b^2} = -1 \Rightarrow a^2 b^2 = 1 \;\; \ldots (i) \)
\( \frac{(a+b)^2}{4} \cdot \frac{\left(\frac{1}{b}-\frac{1}{a}\right)^2}{4} = 1 \)
\( (a+b)^2 (a-b)^2 = 16 a^2 b^2 \)
\( = 16 \Rightarrow (a^2-b^2)^2 = 16 \)
If \( a^2-b^2 = 4 \)
\( \Rightarrow a^2 - \frac{1}{a^2} = 4 \)
\( \Rightarrow a^4 - 4a^2 - 1 = 0 \)
\( \Rightarrow a^2 = \frac{4 \pm \sqrt{20}}{2} \)
\( \Rightarrow a^2 = 2 + \sqrt{5} \; ; \; b^2 = \frac{1}{2+\sqrt{5}} = \sqrt{5} - 2 \)
\( AB^2 = (a-b)^2 + \left(\frac{1}{b} + \frac{1}{a}\right)^2 \)
\( = a^2 + b^2 - 2ab + \frac{1}{a^2} + \frac{1}{b^2} + \frac{2}{ab} \)
\( = 2 + \sqrt{5} + \sqrt{5} - 2 + \sqrt{5} - 2 + \sqrt{5} + 2 \)
\( = 4\sqrt{5} \)
If \( a^2 - b^2 = -4 \)
\( \Rightarrow a^2 - \frac{1}{a^2} + 4 = 0 \)
\( a^4 + 4a^2 - 1 = 0 \)
\( a^2 = \frac{-4 \pm \sqrt{16 + 4}}{2} \)
\( a^2 = -2 \pm \sqrt{5} \)
\( a^2 = \sqrt{5} - 2 \)
\( b^2 = \sqrt{5} + 2 \)
\( AB^2 = (a-b)^2 + \left(\frac{1}{b} + \frac{1}{a}\right)^2 \)
\( = a^2 + b^2 - 2ab + \frac{1}{a^2} + \frac{1}{b^2} + \frac{2}{ab} \)
\( = \sqrt{5} - 2 + \sqrt{5} + 2 + \sqrt{5} + 2 + \sqrt{5} - 2 \)
\( = 4\sqrt{5} \)
Area of square \( = 4\sqrt{5} \)
Its square \( = 80 \)

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