Algebra 2 Worksheets | Conic Sections Worksheets - Free Printable

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

1. Vertical Transverse Axis:
\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\]
- Vertices: $(h, k \pm a)$
- Foci: $(h, k \pm c)$, where $c = \sqrt{a^2 + b^2}$
- Opens: Vertically

2. Horizontal Transverse Axis:
\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]
- Vertices: $(h \pm a, k)$
- Foci: $(h \pm c, k)$, where $c = \sqrt{a^2 + b^2}$
- Opens: Horizontally

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 transverse axis
- Center: $(h, k) = (-6, -3)$
- $a^2 = 16 \implies a = 4$
- $b^2 = 4 \implies b = 2$
- Vertices: $(h, k \pm a) = (-6, -3 \pm 4) = (-6, 1)$ and $(-6, -7)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$
Foci: $(h, k \pm c) = (-6, -3 \pm 2\sqrt{5})$
- Opens: Vertically

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

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

- Type: Vertical transverse axis
- Center: $(h, k) = (-8, 4)$
- $a^2 = 25 \implies a = 5$
- $b^2 = 4 \implies b = 2$
- Vertices: $(h, k \pm a) = (-8, 4 \pm 5) = (-8, 9)$ and $(-8, -1)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{25 + 4} = \sqrt{29}$
Foci: $(h, k \pm c) = (-8, 4 \pm \sqrt{29})$
- Opens: Vertically

Answer:
- Vertices: $(-8, 9)$ and $(-8, -1)$
- Foci: $(-8, 4 + \sqrt{29})$ and $(-8, 4 - \sqrt{29})$
- Opens: Vertically

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

- Type: Vertical transverse axis
- Center: $(h, k) = (4, 3)$
- $a^2 = 121 \implies a = 11$
- $b^2 = 16 \implies b = 4$
- Vertices: $(h, k \pm a) = (4, 3 \pm 11) = (4, 14)$ and $(4, -8)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{121 + 16} = \sqrt{137}$
Foci: $(h, k \pm c) = (4, 3 \pm \sqrt{137})$
- Opens: Vertically

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

---

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

- Type: Vertical transverse axis
- Center: $(h, k) = (0, 4)$
- $a^2 = 81 \implies a = 9$
- $b^2 = 64 \implies b = 8$
- Vertices: $(h, k \pm a) = (0, 4 \pm 9) = (0, 13)$ and $(0, -5)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{81 + 64} = \sqrt{145}$
Foci: $(h, k \pm c) = (0, 4 \pm \sqrt{145})$
- Opens: Vertically

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

---

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

- Type: Horizontal transverse axis
- Center: $(h, k) = (-3, 4)$
- $a^2 = 36 \implies a = 6$
- $b^2 = 9 \implies b = 3$
- Vertices: $(h \pm a, k) = (-3 \pm 6, 4) = (3, 4)$ and $(-9, 4)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$
Foci: $(h \pm c, k) = (-3 \pm 3\sqrt{5}, 4)$
- Opens: Horizontally

Answer:
- Vertices: $(3, 4)$ and $(-9, 4)$
- Foci: $(-3 + 3\sqrt{5}, 4)$ and $(-3 - 3\sqrt{5}, 4)$
- Opens: Horizontally

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

- Type: Horizontal transverse axis
- Center: $(h, k) = (-5, -7)$
- $a^2 = 81 \implies a = 9$
- $b^2 = 49 \implies b = 7$
- Vertices: $(h \pm a, k) = (-5 \pm 9, -7) = (4, -7)$ and $(-14, -7)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{81 + 49} = \sqrt{130}$
Foci: $(h \pm c, k) = (-5 \pm \sqrt{130}, -7)$
- Opens: Horizontally

Answer:
- Vertices: $(4, -7)$ and $(-14, -7)$
- Foci: $(-5 + \sqrt{130}, -7)$ and $(-5 - \sqrt{130}, -7)$
- Opens: Horizontally

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

- Type: Vertical transverse axis
- Center: $(h, k) = (1, -2)$
- $a^2 = 36 \implies a = 6$
- $b^2 = 4 \implies b = 2$
- Vertices: $(h, k \pm a) = (1, -2 \pm 6) = (1, 4)$ and $(1, -8)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$
Foci: $(h, k \pm c) = (1, -2 \pm 2\sqrt{10})$
- Opens: Vertically

Answer:
- Vertices: $(1, 4)$ and $(1, -8)$
- Foci: $(1, -2 + 2\sqrt{10})$ and $(1, -2 - 2\sqrt{10})$
- Opens: Vertically

---

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

- Type: Horizontal transverse axis
- Center: $(h, k) = (-6, 5)$
- $a^2 = 64 \implies a = 8$
- $b^2 = 49 \implies b = 7$
- Vertices: $(h \pm a, k) = (-6 \pm 8, 5) = (2, 5)$ and $(-14, 5)$
- Foci: $c = \sqrt{a^2 + b^2} = \sqrt{64 + 49} = \sqrt{113}$
Foci: $(h \pm c, k) = (-6 \pm \sqrt{113}, 5)$
- Opens: Horizontally

Answer:
- Vertices: $(2, 5)$ and $(-14, 5)$
- Foci: $(-6 + \sqrt{113}, 5)$ and $(-6 - \sqrt{113}, 5)$
- Opens: Horizontally

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

\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: Vertically} \\
2. & \text{Vertices: } (-8, 9) \text{ and } (-8, -1), \text{ Foci: } (-8, 4 + \sqrt{29}) \text{ and } (-8, 4 - \sqrt{29}), \text{ Opens: Vertically} \\
3. & \text{Vertices: } (4, 14) \text{ and } (4, -8), \text{ Foci: } (4, 3 + \sqrt{137}) \text{ and } (4, 3 - \sqrt{137}), \text{ Opens: Vertically} \\
4. & \text{Vertices: } (0, 13) \text{ and } (0, -5), \text{ Foci: } (0, 4 + \sqrt{145}) \text{ and } (0, 4 - \sqrt{145}), \text{ Opens: Vertically} \\
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: Horizontally} \\
6. & \text{Vertices: } (4, -7) \text{ and } (-14, -7), \text{ Foci: } (-5 + \sqrt{130}, -7) \text{ and } (-5 - \sqrt{130}, -7), \text{ Opens: Horizontally} \\
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: Vertically} \\
8. & \text{Vertices: } (2, 5) \text{ and } (-14, 5), \text{ Foci: } (-6 + \sqrt{113}, 5) \text{ and } (-6 - \sqrt{113}, 5), \text{ Opens: Horizontally} \\
\end{array}
}