Question 1209670
Here's how to find the constant k:

1. **Rearrange the equation:** Combine like terms and move everything to one side:

   4x² + 9y² - 8x + 54y + k - 2x² - 5y² + 12x - 34y = 0
   2x² + 4y² + 4x + 20y + k = 0

2. **Complete the square:**  Group the x and y terms and complete the square for each:

   2(x² + 2x) + 4(y² + 5y) + k = 0
   2(x² + 2x + 1) - 2 + 4(y² + 5y + 25/4) - 25 + k = 0
   2(x + 1)² + 4(y + 5/2)² + k - 27 = 0
   2(x + 1)² + 4(y + 5/2)² = 27 - k

3. **Standard form of an ellipse:** Divide by (27 - k) to get the equation in standard form:

   [2(x + 1)²] / (27 - k) + [4(y + 5/2)²] / (27 - k) = 1
   [(x + 1)²] / [(27 - k)/2] + [(y + 5/2)²] / [(27 - k)/4] = 1

4. **Area of an ellipse:** The area of an ellipse is given by A = πab, where a and b are the semi-major and semi-minor axes. In our case:

   a² = (27 - k)/2  => a = √[(27 - k)/2]
   b² = (27 - k)/4  => b = √[(27 - k)/4] = √[(27 - k)/2]/√2 = a/√2

5. **Solve for k:** We're given that the area is 6π:

   6π = πab
   6 = ab
   6 = √[(27 - k)/2] * √[(27 - k)/4]
   6 = (27 - k)/2√2
   12√2 = 27 - k
   k = 27 - 12√2

Therefore, the constant k is 27 - 12√2.