Question 1209671
Here's how to solve this problem:

1. **Coordinate System:** Place the square in a coordinate system so that its center is at the origin (0,0).  Since the side length is 6, the vertices of the square are at (±3, ±3).

2. **Ellipse Equation:** The general equation of an ellipse centered at the origin is (x²/a²) + (y²/b²) = 1, where 'a' is the semi-major axis and 'b' is the semi-minor axis.

3. **Vertices on the Ellipse:** The vertices of the square lie on the ellipse.  Therefore, the points (±3, ±3) must satisfy the ellipse equation:

   (3²/a²) + (3²/b²) = 1
   9/a² + 9/b² = 1

4. **Point P:**  The point P is such that PC = PD = 8.  Let the coordinates of P be (x, y).  Using the distance formula:

   PC² = (x - 3)² + (y - 3)² = 8² = 64
   PD² = (x + 3)² + (y - 3)² = 8² = 64

5. **Solving for x and y:**  Subtract the two equations:

   [(x - 3)² - (x + 3)²] + [(y - 3)² - (y - 3)²] = 0
   (x² - 6x + 9 - x² - 6x - 9) = 0
   -12x = 0
   x = 0

   Now substitute x = 0 into the equation for PC²:

   (0 - 3)² + (y - 3)² = 64
   9 + (y - 3)² = 64
   (y - 3)² = 55
   y - 3 = ±√55
   y = 3 ± √55

   So, the possible coordinates for P are (0, 3 + √55) and (0, 3 - √55). Since the ellipse is circumscribed around the square, the value 3 - √55 must be between -3 and 3. √55 is approximately 7.4, so 3 - √55 ≈ -4.4, which is outside the range. The y coordinate we want is 3 + √55.

6. **Substituting P into the Ellipse Equation:**  Since P(0, 3 + √55) lies on the ellipse:

   (0²/a²) + ( (3 + √55)² / b² ) = 1
   (9 + 6√55 + 55) / b² = 1
   (64 + 6√55) / b² = 1
   b² = 64 + 6√55

7. **Solving for a²:**  Substitute b² back into the equation from step 3:

   9/a² + 9/(64 + 6√55) = 1
   9/a² = 1 - 9/(64 + 6√55)
   9/a² = (64 + 6√55 - 9) / (64 + 6√55)
   9/a² = (55 + 6√55) / (64 + 6√55)
   a² = 9 * (64 + 6√55) / (55 + 6√55)
   a² = 9 * (64 + 6√55) * (55 - 6√55) / (55² - (6√55)²)
   a² = 9 * (3520 - 396√55 + 330√55 - 1980) / (3025 - 1980)
   a² = 9 * 1540 / 1045
   a² = 9 * (28/19) = 252/19

8. **Area of the Ellipse:**  Area = πab = π√(a²b²)

   Area = π√[ (252/19) * (64 + 6√55) ]
   Area = π√[ (252/19) * (64 + 6√55) ] ≈ π√(252/19 * 108.7) ≈ π√1440 ≈ 12π√10 ≈ 119.38

Final Answer: The final answer is $\boxed{12\pi\sqrt{10}}$