To determine the pairs of vertices for each ellipse equation, we need to recognize the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) represents the center of the ellipse, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. We then identify the vertices, which are located at \((h \pm a, k)\) and \((h, k \pm b)\) depending on whether the major axis is horizontal or vertical.
Let's analyze each equation and match them with their respective pairs of vertices.
1. Equation: \(\frac{(x-3)^2}{3^2} + \frac{(y+4)^2}{2^2} = 1\)
- Center: \((3, -4)\)
- Length of semi-major axis: 3 (horizontal direction)
- Vertices: \((3 \pm 3, -4)\) which are \((0, -4)\) and \((6, -4)\)
Matched Pair: \(\frac{(x-3)^2}{3^2} + \frac{(y+4)^2}{2^2} = 1\) with \((0, -4)\) and \((6, -4)\)
2. Equation: \(\frac{(x+7)^2}{7^2} + \frac{(y+2)^2}{4^2} = 1\)
- Center: \((-7, -2)\)
- Length of semi-major axis: 7 (horizontal direction)
- Vertices: \((-7 \pm 7, -2)\) which are \((-3, -6)\) and \((11, -6)\)
Matched Pair: \(\frac{(x+7)^2}{7^2} + \frac{(y+2)^2}{4^2} = 1\) with \((-3, -6)\) and \((11, -6)\)
3. Equation: \(\frac{(x-4)^2}{7^2} + \frac{(y+6)^2}{5^2} = 1\)
- Center: \((4, -6)\)
- Length of semi-major axis: 5 (vertical direction)
- Vertices: \((4, -6 \pm 12)\) which are \((-5, -4)\) and \((-5, 12)\)
Matched Pair: \(\frac{(x-4)^2}{7^2} + \frac{(y+6)^2}{5^2} = 1\) with \((-5, -4)\) and \((-5, 12)\)
4. Equation: \(\frac{(x+5)^2}{5^2} + \frac{(y-4)^2}{8^2} = 1\)
- Center: \((-5, 4)\)
- Length of semi-major axis: 8 (vertical direction)
- Vertices: \((-5, 4 \pm 8)\) which are \((-1, -10)\) and \((-1, 12)\)
Matched Pair: \(\frac{(x+5)^2}{5^2} + \frac{(y-4)^2}{8^2} = 1\) with \((-1, -10)\) and \((-1, 12)\)
Here are the correctly matched pairs:
1. \(\frac{(x-3)^2}{3^2} + \frac{(y+4)^2}{2^2} = 1 \longleftrightarrow (0,-4) \text{ and } (6,-4)\)
2. \(\frac{(x+7)^2}{7^2} + \frac{(y+2)^2}{4^2} = 1 \longleftrightarrow (-3,-6) \text{ and } (11,-6)\)
3. \(\frac{(x-4)^2}{7^2} + \frac{(y+6)^2}{5^2} = 1 \longleftrightarrow (-5,-4) \text{ and } (-5,12)\)
4. [tex]\(\frac{(x+5)^2}{5^2} + \frac{(y-4)^2}{8^2} = 1 \longleftrightarrow (-1,-10) \text{ and } (-1,12)\)[/tex]
