The major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis
To answer the question, we need to write it in the standard form of the equation of an ellipse
The equation of an ellipse centered at (h,k) is
(x - h)²/a² + (y - k)²/b² (1) where a > b and the major axis is parallel to the x axis
Given x² + 16y² - 96y + 128 = 0, we convert it into the standard equation of an ellipse.
So, x² + 16y² - 96y + 128 = 0
Dividing through by 16, we have
x²/16 + 16y²/16 - 96y/16 + 128/16 = 0/16
x²/16 + y² - 6y + 8 = 0
Completing the square in y by adding and subtracting (-6/2)² = (-3)²
x²/16 + y² - 6y + (-3)² - (-3)² + 8 = 0
x²/16 + (y - 3)² - 9 + 8 = 0
x²/16 + (y - 3)² - 1 = 0
x²/16 + (y - 3)² = 1
x²/4² + (y - 3)²/1² = 1 (2)
Comparing equations (1) and (2), we have that a = 4 and b = 1.
Since a = 4 > b = 1, the major axis for the ellipse is the x-axis
So, the major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis
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