SOLUTION: Find the constant k so that the equation 4x^2 + 9y^2 - 8x + 54y + k = 2x^2 + 5y^2 - 12x + 34y represents an ellipse which has an area of 6 \pi.

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Question 1209670: Find the constant k so that the equation
4x^2 + 9y^2 - 8x + 54y + k = 2x^2 + 5y^2 - 12x + 34y
represents an ellipse which has an area of 6 \pi.

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website!
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.

Answer by ikleyn(52802) (Show Source):
You can put this solution on YOUR website!