Question

21. Vertices: \( (2,0),(10,0) \); minor axis of length 4

Ask by Fowler Dawson. in the United States
Mar 18,2025

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The equation of the ellipse is \( \frac{(x-6)^2}{16} + \frac{y^2}{4} = 1 \).

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Bonus Knowledge

To sketch this ellipse, you first need to determine its center and co-vertices. The center of the ellipse is the midpoint of the vertices along the major axis. Here, the vertices are \((2,0)\) and \((10,0)\), which gives a center at \[ \left( \frac{2+10}{2}, \frac{0+0}{2} \right) = (6, 0). \] Next, the semi-major axis \(a\) is half the distance between the vertices, calculated as \[ a = \frac{10 - 2}{2} = 4. \] Since the minor axis is given to be of length 4 as well, the semi-minor axis \(b\) is \[ b = \frac{4}{2} = 2. \] Thus, we have enough information to write the equation of the ellipse as \[ \frac{(x - 6)^2}{4^2} + \frac{(y - 0)^2}{2^2} = 1 \quad \text{or} \quad \frac{(x - 6)^2}{16} + \frac{y^2}{4} = 1. \] Ready to grab your sketching tools? Draw this ellipse centered at (6,0) with your major axis running horizontally and minor axis running vertically. Don't forget to mark the vertices and co-vertices at (2,0), (10,0), (6,2), and (6,-2)!

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