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16. Horizontal major axis; passes through the points \( (5,0) \) and \( (0,2) \) 17. Vertices: \( ( \pm 6,0) \); passes through the point \( (4,1) \) 18. Vertices: \( (0, \pm 8) \); passes through the point \( (3,4) \)

Ask by Turnbull Greene. in the United States
Mar 18,2025

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The equations of the ellipses are: 1. **Problem 16:** \[ \frac{(x - 2.5)^2}{7.25} + \frac{(y - 1)^2}{7.25} = 1 \] 2. **Problem 17:** \[ \frac{x^2}{36} + \frac{y^2}{1.8} = 1 \] 3. **Problem 18:** \[ \frac{x^2}{12} + \frac{y^2}{64} = 1 \]

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The Deep Dive

To find the equations of the conic sections based on the given points, first take note of the vertices given in each scenario. For a hyperbola with a horizontal major axis that passes through the points \( (5,0) \) and \( (0,2) \), you can use the standard form of a hyperbola. The center will be at the midpoint of \( (5,0) \) and the other point, giving a starting point for your equation. For the vertices listed as \( ( \pm 6,0) \), you have an ellipse with a major axis on the x-axis. The equation will start out as \( \frac{x^2}{36} + \frac{y^2}{b^2} = 1 \). You can find \( b \) using the fact that it passes through the point \( (4,1) \) to solve for that missing value. The third scenario with vertices at \( (0, \pm 8) \) suggests a vertical ellipse. Starting with the equation \( \frac{x^2}{a^2} + \frac{y^2}{64} = 1 \), you can plug in the coordinates of the point \( (3,4) \) to determine the value of \( a \) and complete the equation. The mystery of these shapes begins to unravel with a bit of algebra!

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