Math worksheet for identifying properties of hyperbolas.

Let's solve each hyperbola problem step by step and identify the vertices, foci, and direction it opens.

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🔷 General Form of a Hyperbola:

There are two standard forms:

1. Vertical transverse axis (opens up/down):
$$
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
$$
- 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 transverse axis (opens left/right):
$$
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
$$
- Center: $(h, k)$
- Vertices: $(h \pm a, k)$
- Foci: $(h \pm c, k)$, where $c = \sqrt{a^2 + b^2}$
- Opens: Left and right

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We'll go through each one.

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

- This is in the form $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ → vertical
- $k = -3$, $h = -6$ → center: $(-6, -3)$
- $a^2 = 16$ → $a = 4$
- $b^2 = 4$ → $b = 2$
- $c = \sqrt{a^2 + b^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$

✔ Vertices: $(-6, -3 \pm 4)$ → $(-6, 1)$ and $(-6, -7)$
✔ Foci: $(-6, -3 \pm 2\sqrt{5})$
✔ Opens: Up and down

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

- Vertical hyperbola
- Center: $(-8, 4)$
- $a^2 = 25$ → $a = 5$
- $b^2 = 4$ → $b = 2$
- $c = \sqrt{25 + 4} = \sqrt{29}$

✔ Vertices: $(-8, 4 \pm 5)$ → $(-8, 9)$ and $(-8, -1)$
✔ Foci: $(-8, 4 \pm \sqrt{29})$
✔ Opens: Up and down

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

- Vertical
- Center: $(4, 3)$
- $a^2 = 121$ → $a = 11$
- $b^2 = 16$ → $b = 4$
- $c = \sqrt{121 + 16} = \sqrt{137}$

✔ Vertices: $(4, 3 \pm 11)$ → $(4, 14)$ and $(4, -8)$
✔ Foci: $(4, 3 \pm \sqrt{137})$
✔ Opens: Up and down

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

- Note: $x^2$ means $(x - 0)^2$, so center is $(0, 4)$
- Vertical
- $a^2 = 81$ → $a = 9$
- $b^2 = 64$ → $b = 8$
- $c = \sqrt{81 + 64} = \sqrt{145}$

✔ Vertices: $(0, 4 \pm 9)$ → $(0, 13)$ and $(0, -5)$
✔ Foci: $(0, 4 \pm \sqrt{145})$
✔ Opens: Up and down

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

- Horizontal
- Center: $(-3, 4)$
- $a^2 = 36$ → $a = 6$
- $b^2 = 9$ → $b = 3$
- $c = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$

✔ Vertices: $(-3 \pm 6, 4)$ → $(3, 4)$ and $(-9, 4)$
✔ Foci: $(-3 \pm 3\sqrt{5}, 4)$
✔ Opens: Left and right

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

- Horizontal
- Center: $(-5, -7)$
- $a^2 = 81$ → $a = 9$
- $b^2 = 49$ → $b = 7$
- $c = \sqrt{81 + 49} = \sqrt{130}$

✔ Vertices: $(-5 \pm 9, -7)$ → $(4, -7)$ and $(-14, -7)$
✔ Foci: $(-5 \pm \sqrt{130}, -7)$
✔ Opens: Left and right

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

- Vertical
- Center: $(1, -2)$
- $a^2 = 36$ → $a = 6$
- $b^2 = 4$ → $b = 2$
- $c = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$

✔ Vertices: $(1, -2 \pm 6)$ → $(1, 4)$ and $(1, -8)$
✔ Foci: $(1, -2 \pm 2\sqrt{10})$
✔ Opens: Up and down

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

- Horizontal
- Center: $(-6, 5)$
- $a^2 = 64$ → $a = 8$
- $b^2 = 49$ → $b = 7$
- $c = \sqrt{64 + 49} = \sqrt{113}$

✔ Vertices: $(-6 \pm 8, 5)$ → $(2, 5)$ and $(-14, 5)$
✔ Foci: $(-6 \pm \sqrt{113}, 5)$
✔ Opens: Left and right

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✔ Final Answers:

#### 1)
- Vertices: $(-6, 1), (-6, -7)$
- Foci: $(-6, -3 \pm 2\sqrt{5})$
- Opens: Up and down

#### 2)
- Vertices: $(-8, 9), (-8, -1)$
- Foci: $(-8, 4 \pm \sqrt{29})$
- Opens: Up and down

#### 3)
- Vertices: $(4, 14), (4, -8)$
- Foci: $(4, 3 \pm \sqrt{137})$
- Opens: Up and down

#### 4)
- Vertices: $(0, 13), (0, -5)$
- Foci: $(0, 4 \pm \sqrt{145})$
- Opens: Up and down

#### 5)
- Vertices: $(3, 4), (-9, 4)$
- Foci: $(-3 \pm 3\sqrt{5}, 4)$
- Opens: Left and right

#### 6)
- Vertices: $(4, -7), (-14, -7)$
- Foci: $(-5 \pm \sqrt{130}, -7)$
- Opens: Left and right

#### 7)
- Vertices: $(1, 4), (1, -8)$
- Foci: $(1, -2 \pm 2\sqrt{10})$
- Opens: Up and down

#### 8)
- Vertices: $(2, 5), (-14, 5)$
- Foci: $(-6 \pm \sqrt{113}, 5)$
- Opens: Left and right

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Let me know if you'd like this formatted as a printable answer sheet!