5.4 Hyperbola Review.pdf - Duluth High School - Free Printable

Problem Overview:

The worksheet involves identifying and graphing conic sections, specifically hyperbolas and ellipses. The tasks are divided into two parts:
1. Identifying the equations of hyperbolas from their graphs (Questions 1-3).
2. Graphing ellipses and finding their key features (Questions 4-7).

Let's solve each part step by step.

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Part 1: Identifying the Equations of Hyperbolas

#### Question 1:
The graph shows a hyperbola with a vertical transverse axis. The center is at the origin \((0, 0)\), and the vertices are at \((0, \pm 3)\). The asymptotes are also visible.

- Standard Form: For a hyperbola with a vertical transverse axis, the standard form is:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]
- Vertices: The vertices are at \((0, \pm a)\). Here, \(a = 3\), so \(a^2 = 9\).
- Equation: Since the graph does not provide enough information to determine \(b\), we can write the equation as:
\[
\frac{y^2}{9} - \frac{x^2}{b^2} = 1
\]

#### Question 2:
The graph shows a hyperbola with a horizontal transverse axis. The center is at the origin \((0, 0)\), and the vertices are at \((\pm 5, 0)\). The asymptotes are also visible.

- Standard Form: For a hyperbola with a horizontal transverse axis, the standard form is:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\]
- Vertices: The vertices are at \((\pm a, 0)\). Here, \(a = 5\), so \(a^2 = 25\).
- Equation: Since the graph does not provide enough information to determine \(b\), we can write the equation as:
\[
\frac{x^2}{25} - \frac{y^2}{b^2} = 1
\]

#### Question 3:
The graph shows a hyperbola with a vertical transverse axis. The center is at \((0, 0)\), and the vertices are at \((0, \pm 4)\). The asymptotes are also visible.

- Standard Form: For a hyperbola with a vertical transverse axis, the standard form is:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]
- Vertices: The vertices are at \((0, \pm a)\). Here, \(a = 4\), so \(a^2 = 16\).
- Equation: Since the graph does not provide enough information to determine \(b\), we can write the equation as:
\[
\frac{y^2}{16} - \frac{x^2}{b^2} = 1
\]

---

Part 2: Graphing Ellipses and Finding Key Features

#### Question 4:
The given equation is:
\[
\frac{(x + 1)^2}{16} + \frac{(y - 2)^2}{25} = 1
\]

- Center: The center is \((-1, 2)\).
- Vertices: Since the denominator under \((y - 2)^2\) is larger (\(25 > 16\)), the major axis is vertical. The vertices are at \((-1, 2 \pm 5)\), i.e., \((-1, 7)\) and \((-1, -3)\).
- Co-vertices: The co-vertices are at \((-1 \pm 4, 2)\), i.e., \((3, 2)\) and \((-5, 2)\).
- Foci: The distance from the center to each focus is \(c\), where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 25\) and \(b^2 = 16\), so:
\[
c = \sqrt{25 - 16} = \sqrt{9} = 3
\]
The foci are at \((-1, 2 \pm 3)\), i.e., \((-1, 5)\) and \((-1, -1)\).

#### Question 5:
The given equation is:
\[
\frac{(x - 1)^2}{36} + \frac{y^2}{9} = 1
\]

- Center: The center is \((1, 0)\).
- Vertices: Since the denominator under \((x - 1)^2\) is larger (\(36 > 9\)), the major axis is horizontal. The vertices are at \((1 \pm 6, 0)\), i.e., \((7, 0)\) and \((-5, 0)\).
- Co-vertices: The co-vertices are at \((1, \pm 3)\), i.e., \((1, 3)\) and \((1, -3)\).
- Foci: The distance from the center to each focus is \(c\), where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 36\) and \(b^2 = 9\), so:
\[
c = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}
\]
The foci are at \((1 \pm 3\sqrt{3}, 0)\).

#### Question 6:
The given equation is:
\[
\frac{x^2}{36} + \frac{(y + 1)^2}{25} = 1
\]

- Center: The center is \((0, -1)\).
- Vertices: Since the denominator under \((y + 1)^2\) is larger (\(25 > 36\)), the major axis is vertical. The vertices are at \((0, -1 \pm 5)\), i.e., \((0, 4)\) and \((0, -6)\).
- Co-vertices: The co-vertices are at \((\pm 6, -1)\), i.e., \((6, -1)\) and \((-6, -1)\).
- Foci: The distance from the center to each focus is \(c\), where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 25\) and \(b^2 = 36\), so:
\[
c = \sqrt{36 - 25} = \sqrt{11}
\]
The foci are at \((0, -1 \pm \sqrt{11})\).

#### Question 7:
The given equation is:
\[
\frac{(x + 1)^2}{16} + \frac{(y - 2)^2}{36} = 1
\]

- Center: The center is \((-1, 2)\).
- Vertices: Since the denominator under \((y - 2)^2\) is larger (\(36 > 16\)), the major axis is vertical. The vertices are at \((-1, 2 \pm 6)\), i.e., \((-1, 8)\) and \((-1, -4)\).
- Co-vertices: The co-vertices are at \((-1 \pm 4, 2)\), i.e., \((3, 2)\) and \((-5, 2)\).
- Foci: The distance from the center to each focus is \(c\), where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 36\) and \(b^2 = 16\), so:
\[
c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}
\]
The foci are at \((-1, 2 \pm 2\sqrt{5})\).

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Final Answers:

1. \(\frac{y^2}{9} - \frac{x^2}{b^2} = 1\)
2. \(\frac{x^2}{25} - \frac{y^2}{b^2} = 1\)
3. \(\frac{y^2}{16} - \frac{x^2}{b^2} = 1\)

4. Center: \((-1, 2)\), Vertices: \((-1, 7)\) and \((-1, -3)\), Co-vertices: \((3, 2)\) and \((-5, 2)\), Foci: \((-1, 5)\) and \((-1, -1)\)

5. Center: \((1, 0)\), Vertices: \((7, 0)\) and \((-5, 0)\), Co-vertices: \((1, 3)\) and \((1, -3)\), Foci: \((1 + 3\sqrt{3}, 0)\) and \((1 - 3\sqrt{3}, 0)\)

6. Center: \((0, -1)\), Vertices: \((0, 4)\) and \((0, -6)\), Co-vertices: \((6, -1)\) and \((-6, -1)\), Foci: \((0, -1 + \sqrt{11})\) and \((0, -1 - \sqrt{11})\)

7. Center: \((-1, 2)\), Vertices: \((-1, 8)\) and \((-1, -4)\), Co-vertices: \((3, 2)\) and \((-5, 2)\), Foci: \((-1, 2 + 2\sqrt{5})\) and \((-1, 2 - 2\sqrt{5})\)

\boxed{
\begin{aligned}
&1. \frac{y^2}{9} - \frac{x^2}{b^2} = 1 \\
&2. \frac{x^2}{25} - \frac{y^2}{b^2} = 1 \\
&3. \frac{y^2}{16} - \frac{x^2}{b^2} = 1 \\
&4. \text{Center: } (-1, 2), \text{ Vertices: } (-1, 7) \text{ and } (-1, -3), \text{ Co-vertices: } (3, 2) \text{ and } (-5, 2), \text{ Foci: } (-1, 5) \text{ and } (-1, -1) \\
&5. \text{Center: } (1, 0), \text{ Vertices: } (7, 0) \text{ and } (-5, 0), \text{ Co-vertices: } (1, 3) \text{ and } (1, -3), \text{ Foci: } (1 + 3\sqrt{3}, 0) \text{ and } (1 - 3\sqrt{3}, 0) \\
&6. \text{Center: } (0, -1), \text{ Vertices: } (0, 4) \text{ and } (0, -6), \text{ Co-vertices: } (6, -1) \text{ and } (-6, -1), \text{ Foci: } (0, -1 + \sqrt{11}) \text{ and } (0, -1 - \sqrt{11}) \\
&7. \text{Center: } (-1, 2), \text{ Vertices: } (-1, 8) \text{ and } (-1, -4), \text{ Co-vertices: } (3, 2) \text{ and } (-5, 2), \text{ Foci: } (-1, 2 + 2\sqrt{5}) \text{ and } (-1, 2 - 2\sqrt{5})
\end{aligned}
}