Identify properties of hyperbolas from given equations in this math worksheet.

To solve the problem of identifying the properties of hyperbolas, we need to analyze each given equation step by step. The general forms of hyperbolas are:

1. Vertical Hyperbola: \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\)
- Center: \((h, k)\)
- Vertices: \((h, k \pm a)\)
- Foci: \((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\)
- Opens: Up and down

2. Horizontal Hyperbola: \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\)
- Center: \((h, k)\)
- Vertices: \((h \pm a, k)\)
- Foci: \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\)
- Opens: Left and right

Let's solve each equation one by one.

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1) \(\frac{(y + 3)^2}{16} - \frac{(x + 6)^2}{4} = 1\)

- Type: Vertical hyperbola
- Center: \((-6, -3)\)
- \(a^2 = 16 \Rightarrow a = 4\), so vertices are \((-6, -3 \pm 4)\)
- Vertices: \((-6, 1)\) and \((-6, -7)\)
- \(b^2 = 4 \Rightarrow b = 2\), so \(c = \sqrt{a^2 + b^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\)
- Foci: \((-6, -3 \pm 2\sqrt{5})\)
- Opens: Up and down

Answer for 1):
- Vertices: \((-6, 1)\) and \((-6, -7)\)
- Foci: \((-6, -3 + 2\sqrt{5})\) and \((-6, -3 - 2\sqrt{5})\)
- Opens: Up and down

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2) \(\frac{(y - 4)^2}{25} - \frac{(x + 8)^2}{4} = 1\)

- Type: Vertical hyperbola
- Center: \((-8, 4)\)
- \(a^2 = 25 \Rightarrow a = 5\), so vertices are \((-8, 4 \pm 5)\)
- Vertices: \((-8, 9)\) and \((-8, -1)\)
- \(b^2 = 4 \Rightarrow b = 2\), so \(c = \sqrt{a^2 + b^2} = \sqrt{25 + 4} = \sqrt{29}\)
- Foci: \((-8, 4 \pm \sqrt{29})\)
- Opens: Up and down

Answer for 2):
- Vertices: \((-8, 9)\) and \((-8, -1)\)
- Foci: \((-8, 4 + \sqrt{29})\) and \((-8, 4 - \sqrt{29})\)
- Opens: Up and down

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3) \(\frac{(y - 3)^2}{121} - \frac{(x - 4)^2}{16} = 1\)

- Type: Vertical hyperbola
- Center: \((4, 3)\)
- \(a^2 = 121 \Rightarrow a = 11\), so vertices are \((4, 3 \pm 11)\)
- Vertices: \((4, 14)\) and \((4, -8)\)
- \(b^2 = 16 \Rightarrow b = 4\), so \(c = \sqrt{a^2 + b^2} = \sqrt{121 + 16} = \sqrt{137}\)
- Foci: \((4, 3 \pm \sqrt{137})\)
- Opens: Up and down

Answer for 3):
- Vertices: \((4, 14)\) and \((4, -8)\)
- Foci: \((4, 3 + \sqrt{137})\) and \((4, 3 - \sqrt{137})\)
- Opens: Up and down

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4) \(\frac{(y - 4)^2}{81} - \frac{x^2}{64} = 1\)

- Type: Vertical hyperbola
- Center: \((0, 4)\)
- \(a^2 = 81 \Rightarrow a = 9\), so vertices are \((0, 4 \pm 9)\)
- Vertices: \((0, 13)\) and \((0, -5)\)
- \(b^2 = 64 \Rightarrow b = 8\), so \(c = \sqrt{a^2 + b^2} = \sqrt{81 + 64} = \sqrt{145}\)
- Foci: \((0, 4 \pm \sqrt{145})\)
- Opens: Up and down

Answer for 4):
- Vertices: \((0, 13)\) and \((0, -5)\)
- Foci: \((0, 4 + \sqrt{145})\) and \((0, 4 - \sqrt{145})\)
- Opens: Up and down

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5) \(\frac{(x + 3)^2}{36} - \frac{(y - 4)^2}{9} = 1\)

- Type: Horizontal hyperbola
- Center: \((-3, 4)\)
- \(a^2 = 36 \Rightarrow a = 6\), so vertices are \((-3 \pm 6, 4)\)
- Vertices: \((3, 4)\) and \((-9, 4)\)
- \(b^2 = 9 \Rightarrow b = 3\), so \(c = \sqrt{a^2 + b^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\)
- Foci: \((-3 \pm 3\sqrt{5}, 4)\)
- Opens: Left and right

Answer for 5):
- Vertices: \((3, 4)\) and \((-9, 4)\)
- Foci: \((-3 + 3\sqrt{5}, 4)\) and \((-3 - 3\sqrt{5}, 4)\)
- Opens: Left and right

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6) \(\frac{(x + 5)^2}{81} - \frac{(y + 7)^2}{49} = 1\)

- Type: Horizontal hyperbola
- Center: \((-5, -7)\)
- \(a^2 = 81 \Rightarrow a = 9\), so vertices are \((-5 \pm 9, -7)\)
- Vertices: \((4, -7)\) and \((-14, -7)\)
- \(b^2 = 49 \Rightarrow b = 7\), so \(c = \sqrt{a^2 + b^2} = \sqrt{81 + 49} = \sqrt{130}\)
- Foci: \((-5 \pm \sqrt{130}, -7)\)
- Opens: Left and right

Answer for 6):
- Vertices: \((4, -7)\) and \((-14, -7)\)
- Foci: \((-5 + \sqrt{130}, -7)\) and \((-5 - \sqrt{130}, -7)\)
- Opens: Left and right

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7) \(\frac{(y + 2)^2}{36} - \frac{(x - 1)^2}{4} = 1\)

- Type: Vertical hyperbola
- Center: \((1, -2)\)
- \(a^2 = 36 \Rightarrow a = 6\), so vertices are \((1, -2 \pm 6)\)
- Vertices: \((1, 4)\) and \((1, -8)\)
- \(b^2 = 4 \Rightarrow b = 2\), so \(c = \sqrt{a^2 + b^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\)
- Foci: \((1, -2 \pm 2\sqrt{10})\)
- Opens: Up and down

Answer for 7):
- Vertices: \((1, 4)\) and \((1, -8)\)
- Foci: \((1, -2 + 2\sqrt{10})\) and \((1, -2 - 2\sqrt{10})\)
- Opens: Up and down

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8) \(\frac{(x + 6)^2}{64} - \frac{(y - 5)^2}{49} = 1\)

- Type: Horizontal hyperbola
- Center: \((-6, 5)\)
- \(a^2 = 64 \Rightarrow a = 8\), so vertices are \((-6 \pm 8, 5)\)
- Vertices: \((2, 5)\) and \((-14, 5)\)
- \(b^2 = 49 \Rightarrow b = 7\), so \(c = \sqrt{a^2 + b^2} = \sqrt{64 + 49} = \sqrt{113}\)
- Foci: \((-6 \pm \sqrt{113}, 5)\)
- Opens: Left and right

Answer for 8):
- Vertices: \((2, 5)\) and \((-14, 5)\)
- Foci: \((-6 + \sqrt{113}, 5)\) and \((-6 - \sqrt{113}, 5)\)
- Opens: Left and right

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Final Answers:

1. Vertices: \((-6, 1)\) and \((-6, -7)\), Foci: \((-6, -3 + 2\sqrt{5})\) and \((-6, -3 - 2\sqrt{5})\), Opens: Up and down
2. Vertices: \((-8, 9)\) and \((-8, -1)\), Foci: \((-8, 4 + \sqrt{29})\) and \((-8, 4 - \sqrt{29})\), Opens: Up and down
3. Vertices: \((4, 14)\) and \((4, -8)\), Foci: \((4, 3 + \sqrt{137})\) and \((4, 3 - \sqrt{137})\), Opens: Up and down
4. Vertices: \((0, 13)\) and \((0, -5)\), Foci: \((0, 4 + \sqrt{145})\) and \((0, 4 - \sqrt{145})\), Opens: Up and down
5. Vertices: \((3, 4)\) and \((-9, 4)\), Foci: \((-3 + 3\sqrt{5}, 4)\) and \((-3 - 3\sqrt{5}, 4)\), Opens: Left and right
6. Vertices: \((4, -7)\) and \((-14, -7)\), Foci: \((-5 + \sqrt{130}, -7)\) and \((-5 - \sqrt{130}, -7)\), Opens: Left and right
7. Vertices: \((1, 4)\) and \((1, -8)\), Foci: \((1, -2 + 2\sqrt{10})\) and \((1, -2 - 2\sqrt{10})\), Opens: Up and down
8. Vertices: \((2, 5)\) and \((-14, 5)\), Foci: \((-6 + \sqrt{113}, 5)\) and \((-6 - \sqrt{113}, 5)\), Opens: Left and right

\[
\boxed{
\begin{array}{ll}
1. & \text{Vertices: } (-6, 1) \text{ and } (-6, -7), \text{ Foci: } (-6, -3 + 2\sqrt{5}) \text{ and } (-6, -3 - 2\sqrt{5}), \text{ Opens: Up and down} \\
2. & \text{Vertices: } (-8, 9) \text{ and } (-8, -1), \text{ Foci: } (-8, 4 + \sqrt{29}) \text{ and } (-8, 4 - \sqrt{29}), \text{ Opens: Up and down} \\
3. & \text{Vertices: } (4, 14) \text{ and } (4, -8), \text{ Foci: } (4, 3 + \sqrt{137}) \text{ and } (4, 3 - \sqrt{137}), \text{ Opens: Up and down} \\
4. & \text{Vertices: } (0, 13) \text{ and } (0, -5), \text{ Foci: } (0, 4 + \sqrt{145}) \text{ and } (0, 4 - \sqrt{145}), \text{ Opens: Up and down} \\
5. & \text{Vertices: } (3, 4) \text{ and } (-9, 4), \text{ Foci: } (-3 + 3\sqrt{5}, 4) \text{ and } (-3 - 3\sqrt{5}, 4), \text{ Opens: Left and right} \\
6. & \text{Vertices: } (4, -7) \text{ and } (-14, -7), \text{ Foci: } (-5 + \sqrt{130}, -7) \text{ and } (-5 - \sqrt{130}, -7), \text{ Opens: Left and right} \\
7. & \text{Vertices: } (1, 4) \text{ and } (1, -8), \text{ Foci: } (1, -2 + 2\sqrt{10}) \text{ and } (1, -2 - 2\sqrt{10}), \text{ Opens: Up and down} \\
8. & \text{Vertices: } (2, 5) \text{ and } (-14, 5), \text{ Foci: } (-6 + \sqrt{113}, 5) \text{ and } (-6 - \sqrt{113}, 5), \text{ Opens: Left and right} \\
\end{array}
}
\]