Question

Find the center of the ellipse given by the equation \[ 4x^2 + 24x + 9y^2 - 18y + 9 = 0 \]

• A. \( (h, k) = (-3, 1) \)
• B. \( (h, k) = (3, -1) \)
• C. \( (h, k) = (-2, 3) \)
• D. \( (h, k) = (0, 0) \)

Hint:

To find the center of an ellipse, rewrite the equation in standard form by completing the square for both the \( x \) and \( y \) terms. The center will be \( (h, k) \), where the equation is of the form \( \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \).

Answer 1

The Correct Option is A

We are given the equation of the ellipse: \[ 4x^2 + 24x + 9y^2 - 18y + 9 = 0 \] ### Step 1: Group the \( x \) and \( y \) Terms Rearrange the equation: \[ 4(x^2 + 6x) + 9(y^2 - 2y) = -9 \] ### Step 2: Complete the Square for the \( x \)-terms For the \( x \)-terms, \( x^2 + 6x \), the coefficient of \( x \) is 6. To complete the square, take half of 6, which is 3, and square it to get 9. Add and subtract 9 inside the parentheses: \[ 4(x^2 + 6x + 9 - 9) + 9(y^2 - 2y) = -9 \] This becomes: \[ 4((x + 3)^2 - 9) + 9(y^2 - 2y) = -9 \] ### Step 3: Complete the Square for the \( y \)-terms For the \( y \)-terms, \( y^2 - 2y \), the coefficient of \( y \) is -2. To complete the square, take half of -2, which is -1, and square it to get 1. Add and subtract 1 inside the parentheses: \[ 4((x + 3)^2 - 9) + 9((y - 1)^2 - 1) = -9 \] ### Step 4: Simplify the Equation Now, simplify the equation: \[ 4(x + 3)^2 - 36 + 9(y - 1)^2 - 9 = -9 \] \[ 4(x + 3)^2 + 9(y - 1)^2 - 45 = -9 \] \[ 4(x + 3)^2 + 9(y - 1)^2 = 36 \] ### Step 5: Identify the Center The equation is now in standard form for an ellipse: \[ \frac{(x + 3)^2}{9} + \frac{(y - 1)^2}{4} = 1 \] From this, we see that the center of the ellipse is at \( (h, k) = (-3, 1) \). Thus, the correct answer is: \[ \boxed{(A) (h, k) = (-3, 1)} \]