Four points lying on the curves `x^2y^2=1` form a square such that mid-points os sides also lies on the given curve. Find the square of area of the square

Answer

To solve the problem, we need to find the square of the area of a square formed by four points lying on the curve \(x^2y^2 = 1\) such that the midpoints of the sides of the square also lie on the curve.

### Step-by-Step Solution:

1. **Understanding the Curve**:
The equation \(x^2y^2 = 1\) can be rewritten as \(xy = \pm 1\). This represents two rectangular hyperbolas: one in the first and third quadrants (where \(xy = 1\)) and the other in the second and fourth quadrants (where \(xy = -1\)).

2. **Identifying Points on the Curve**:
Let the points on the curve be \(P(\alpha, \frac{1}{\alpha})\), \(Q(\beta, -\frac{1}{\beta})\), \(R(-\alpha, -\frac{1}{\alpha})\), and \(S(-\beta, \frac{1}{\beta})\). These points are symmetric about the origin.

3. **Finding Midpoints**:
The midpoints of the sides of the square can be calculated as follows:
- Midpoint of \(PQ\): \(M_1 = \left(\frac{\alpha + \beta}{2}, \frac{\frac{1}{\alpha} - \frac{1}{\beta}}{2}\right)\)
- Midpoint of \(QR\): \(M_2 = \left(\frac{-\alpha - \beta}{2}, \frac{-\frac{1}{\alpha} - \frac{1}{\beta}}{2}\right)\)
- Midpoint of \(RS\): \(M_3 = \left(\frac{-\alpha + \beta}{2}, \frac{-\frac{1}{\alpha} + \frac{1}{\beta}}{2}\right)\)
- Midpoint of \(SP\): \(M_4 = \left(\frac{\alpha - \beta}{2}, \frac{\frac{1}{\alpha} + \frac{1}{\beta}}{2}\right)\)

4. **Condition for Midpoints**:
Each midpoint must also satisfy the curve equation \(x^2y^2 = 1\). Thus, we need to ensure that:
\[ \left(\frac{\alpha + \beta}{2}\right)^2 \left(\frac{\frac{1}{\alpha} - \frac{1}{\beta}}{2}\right)^2 = 1 \]

5. **Calculating Area**:
The area \(A\) of the square can be calculated using the distance between two adjacent points (say \(P\) and \(Q\)):
\[ A = OP \cdot OQ = \sqrt{\alpha^2 + \left(\frac{1}{\alpha}\right)^2} \cdot \sqrt{\beta^2 + \left(-\frac{1}{\beta}\right)^2} \]
Simplifying this gives:
\[ A = \sqrt{\alpha^2 + \frac{1}{\alpha^2}} \cdot \sqrt{\beta^2 + \frac{1}{\beta^2}} \]

6. **Final Area Calculation**:
The area of the square is \(4A\), and thus the square of the area is:
\[ \text{Area}^2 = (4A)^2 = 16A^2 \]

7. **Substituting Values**:
After substituting the values of \(\alpha^2\) and \(\beta^2\) derived from the conditions, we find that the square of the area of the square is \(80\).

### Final Answer:
The square of the area of the square is \(80\).