Question

The centre of the ellipse $4x^{2}+24x+9y^{2}-18y+9=0$ is ________.

• A. $(1, 3)$
• B. $(-1, 3)$
• C. $(3, 1)$
• D. $(-3, 1)$
• E. $(3, 3)$

Hint:

The centre is found by setting the partial derivatives with respect to $x$ and $y$ to zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The given equation is the general form of an ellipse. To find its center, we need to convert this equation into the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center. This is done by the method of completing the square.
Step 2: Key Formula or Approach
1. Group the x-terms and y-terms together.
2. Factor out the coefficients of \(x^2\) and \(y^2\).
3. Complete the square for both the x and y expressions. Remember to balance the equation by adding the same values to the other side.
4. Rearrange the equation into standard form to identify the center \((h, k)\).
Step 3: Detailed Explanation
1. Group the terms.
The given equation is \(4x^2 + 24x + 9y^2 - 18y + 9 = 0\).
\[ (4x^2 + 24x) + (9y^2 - 18y) + 9 = 0 \] 2. Factor out coefficients.
\[ 4(x^2 + 6x) + 9(y^2 - 2y) + 9 = 0 \] 3. Complete the square.
- For the x-terms: Take half of the coefficient of x (which is 6), square it (\(3^2 = 9\)), and add it inside the parenthesis. We add \(4 \times 9 = 36\) to the equation.
\[ 4(x^2 + 6x + 9) \] - For the y-terms: Take half of the coefficient of y (which is -2), square it (\((-1)^2 = 1\)), and add it inside the parenthesis. We add \(9 \times 1 = 9\) to the equation.
\[ 9(y^2 - 2y + 1) \] Now, modify the original equation. To keep it balanced, we add the same amounts to the right side.
\[ 4(x^2 + 6x + 9) + 9(y^2 - 2y + 1) + 9 = 36 + 9 \] \[ 4(x+3)^2 + 9(y-1)^2 + 9 = 45 \] 4. Rearrange into standard form.
\[ 4(x+3)^2 + 9(y-1)^2 = 45 - 9 \] \[ 4(x+3)^2 + 9(y-1)^2 = 36 \] Divide the entire equation by 36 to make the right side equal to 1.
\[ \frac{4(x+3)^2}{36} + \frac{9(y-1)^2}{36} = \frac{36}{36} \] \[ \frac{(x+3)^2}{9} + \frac{(y-1)^2}{4} = 1 \] 5. Identify the center.
Comparing this with the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we have:
\(x-h = x+3 \implies h = -3\)
\(y-k = y-1 \implies k = 1\)
The center \((h, k)\) is (-3, 1).
Step 4: Final Answer
The centre of the ellipse is (-3, 1).