Question

The center of the ellipse \( 4x^2 + y^2 - 8x + 4y - 8 = 0 \) is:

• A. \( (0, 2) \)
• B. \( (2, -1) \)
• C. \( (2, 1) \)
• D. \( (1, 2) \)
• E. \( (1, -2) \)

Hint:

A faster way to find the center \( (h, k) \) of a conic \( f(x, y) = 0 \) is to solve the partial derivatives: \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \). Here, \( 8x - 8 = 0 \implies x = 1 \) and \( 2y + 4 = 0 \implies y = -2 \).

Answer 1

Answer: (E)

Concept: The standard equation of an ellipse is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \( (h, k) \) is the center of the ellipse. To find the center from a general equation, we use the process of completing the square for both the \( x \) and \( y \) variables. This groups the linear and quadratic terms into squared binomials.

Step 1: Group terms and prepare for completing the square.
Rearrange the equation by grouping \( x \) and \( y \) terms together and moving the constant to the other side: \[ (4x^2 - 8x) + (y^2 + 4y) = 8 \] Factor out the leading coefficient from the \( x \) group: \[ 4(x^2 - 2x) + (y^2 + 4y) = 8 \]

Step 2: Complete the squares and identify the center.
To complete the square for \( x^2 - 2x \), add \( (\frac{-2}{2})^2 = 1 \) inside the parenthesis. To complete the square for \( y^2 + 4y \), add \( (\frac{4}{2})^2 = 4 \). Remember to add the corresponding values to the right side of the equation to maintain equality: \[ 4(x^2 - 2x + 1) + (y^2 + 4y + 4) = 8 + 4(1) + 4 \] \[ 4(x - 1)^2 + (y + 2)^2 = 16 \] Rewriting in the standard form \( \frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{16} = 1 \). Comparing this to \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), we identify: \( h = 1 \) and \( k = -2 \). The center is \( (1, -2) \).