Final answer:
The major axis of the ellipse is 8 units, the minor axis is 4 units. The vertices are located at (8, 0) and (-8, 0), and the foci are located at (3.464, 0) and (-3.464, 0). The eccentricity of the ellipse is approximately 0.433, indicating that it is elongated but still fairly close to a circle.
Step-by-step explanation:
The given equation is x² + 4y² = 16. To find the major and minor axes of the ellipse, we need to rewrite the equation in standard form. Dividing both sides by 16, we get x²/16 + y²/4 = 1. Comparing this to the standard form (x²/a² + y²/b² = 1) of an ellipse, we can see that a² = 16 and b² = 4. Therefore, the length of the major axis is 2a = 2√16 = 8 units, and the length of the minor axis is 2b = 2√4 = 4 units.
Since a² > b², the ellipse is longer horizontally and narrower vertically. The center of the ellipse is at (0,0), and the foci can be found using the formula c = √(a² - b²). So, c = √(16 - 4) = √12 ≈ 3.464 units.
The vertices are located at (±a, 0), so the vertices of the ellipse are (8, 0) and (-8, 0). The foci are located at (±c, 0), so the foci of the ellipse are (3.464, 0) and (-3.464, 0). The eccentricity of the ellipse can be calculated using the formula e = c/a, where c is the distance from the center to a focus, and a is half the length of the major axis.
Therefore, the eccentricity of this ellipse is e = 3.464/8 ≈ 0.433. Since the eccentricity is less than 1, this ellipse can be described as elongated but still fairly close to a circle.