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
<h3>Equation of an ellipse</h3>
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
Learn more about ellipse here:
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![b=\sqrt[3]{\frac{95}{4}}](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D)
![y=4(\sqrt[3]{\frac{95}{4}})^{x}](https://tex.z-dn.net/?f=y%3D4%28%5Csqrt%5B3%5D%7B%5Cfrac%7B95%7D%7B4%7D%7D%29%5E%7Bx%7D)
