Algebra 2 Summer Packet - Free Printable

Let’s solve each hyperbola problem step by step. We’ll find the Vertices, Foci, and which direction it Opens for each equation.

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🔁 Quick Reminder (for you, not to be shown if ExplainAnswer is false):

For a hyperbola in standard form:

- If it looks like:
$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ → opens up/down
Center: $(h, k)$
Vertices: $(h, k \pm a)$
Foci: $(h, k \pm c)$, where $c = \sqrt{a^2 + b^2}$

- If it looks like:
$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ → opens left/right
Center: $(h, k)$
Vertices: $(h \pm a, k)$
Foci: $(h \pm c, k)$, where $c = \sqrt{a^2 + b^2}$

We’ll apply this to each problem.

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

→ This is vertical hyperbola (y-term first).

Center: $(-6, -3)$
$a^2 = 16 → a = 4$
$b^2 = 4 → b = 2$
$c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$

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

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## Problem 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: $(-8, 4 \pm 5) → (-8, 9), (-8, -1)$
Foci: $(-8, 4 \pm \sqrt{29})$
Opens: Up and Down

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

Vertical hyperbola.

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

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

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

Note: $x^2 = (x - 0)^2$, so center x = 0.

Vertical hyperbola.

Center: $(0, 4)$
$a^2 = 81 → a = 9$
$b^2 = 64 → b = 8$
$c = \sqrt{81 + 64} = \sqrt{145}$

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

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

Horizontal hyperbola (x-term first).

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

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

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

Horizontal hyperbola.

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

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

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

Vertical hyperbola.

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

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

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

Horizontal hyperbola.

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

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

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Final Answer:

1)
Vertices: (-6, 1), (-6, -7)
Foci: (-6, -3 ± 2√5)
Opens: Up and Down

2)
Vertices: (-8, 9), (-8, -1)
Foci: (-8, 4 ± √29)
Opens: Up and Down

3)
Vertices: (4, 14), (4, -8)
Foci: (4, 3 ± √137)
Opens: Up and Down

4)
Vertices: (0, 13), (0, -5)
Foci: (0, 4 ± √145)
Opens: Up and Down

5)
Vertices: (3, 4), (-9, 4)
Foci: (-3 ± 3√5, 4)
Opens: Left and Right

6)
Vertices: (4, -7), (-14, -7)
Foci: (-5 ± √130, -7)
Opens: Left and Right

7)
Vertices: (1, 4), (1, -8)
Foci: (1, -2 ± 2√10)
Opens: Up and Down

8)
Vertices: (2, 5), (-14, 5)
Foci: (-6 ± √113, 5)
Opens: Left and Right