https://file.aiquickdraw.com/mathbot/file/1725807048984ixhifeug.jpg

Let's analyze and describe the given hyperbolas one by one.

### Hyperbola 1: 
\[
\frac{(y + 1)^2}{36} - \frac{(x - 1)^2}{64} = 1
\]

#### Key Features:
1. **Orientation:** Since the \(y\)-term is positive and comes first, this is a vertically oriented hyperbola.
2. **Center:** \((h, k) = (1, -1)\)
3. **Vertices:** Since \(a^2 = 36\), \(a = 6\). The vertices are at \((1, -1 \pm 6)\), which means \((1, -7)\) and \((1, 5)\).
4. **Co-vertices:** Since \(b^2 = 64\), \(b = 8\). The co-vertices are at \((1 \pm 8, -1)\), which means \((-7, -1)\) and \((9, -1)\).
5. **Foci:** The focal distance \(c\) is found using \(c^2 = a^2 + b^2 = 36 + 64 = 100\), so \(c = 10\). The foci are at \((1, -1 \pm 10)\), which means \((1, -11)\) and \((1, 9)\).
6. **Length of the Latus Rectum (LLR):** \(2b^2/a = 2 \times 64 / 6 = 21.33\).
7. **Directrices:** Given by the equations \(y = -1 \pm \frac{c^2}{a} = -1 \pm \frac{100}{6}\), which is \(y = -1 \pm 16.67\).
8. **Asymptotes:** The asymptotes are lines through the center with slopes \(\pm \frac{a}{b}\), so they are \(y + 1 = \pm \frac{6}{8}(x - 1)\), simplified as \(y + 1 = \pm \frac{3}{4}(x - 1)\).

### Hyperbola 2:
\[
9x^2 - 4y^2 - 36 = 0
\]

To simplify, rewrite it in standard form by dividing everything by 36:

\[
\frac{9x^2}{36} - \frac{4y^2}{36} = 1 \implies \frac{x^2}{4} - \frac{y^2}{9} = 1
\]

#### Key Features:
1. **Orientation:** Since the \(x\)-term is positive and comes first, this is a horizontally oriented hyperbola.
2. **Center:** \((h, k) = (0, 0)\)
3. **Vertices:** Since \(a^2 = 4\), \(a = 2\). The vertices are at \((\pm 2, 0)\), which means \((-2, 0)\) and \((2, 0)\).
4. **Co-vertices:** Since \(b^2 = 9\), \(b = 3\). The co-vertices are at \((0, \pm 3)\), which means \((0, -3)\) and \((0, 3)\).
5. **Foci:** The focal distance \(c\) is found using \(c^2 = a^2 + b^2 = 4 + 9 = 13\), so \(c = \sqrt{13}\). The foci are at \((\pm \sqrt{13}, 0)\), which is approximately \((\pm 3.6, 0)\).
6. **Length of the Latus Rectum (LLR):** \(2b^2/a = 2 \times 9 / 2 = 9\).
7. **Directrices:** Given by the equations \(x = \pm \frac{c^2}{a} = \pm \frac{13}{2}\), which is approximately \(x = \pm 6.5\).
8. **Asymptotes:** The asymptotes are lines through the center with slopes \(\pm \frac{b}{a}\), so they are \(y = \pm \frac{3}{2}x\).

Would you like more details or have any specific questions?

### Further Questions:
1. How do we determine the orientation of a hyperbola?
2. Can you explain the method to find the foci of a hyperbola?
3. What are the key differences between vertical and horizontal hyperbolas?
4. How are the equations of asymptotes derived for hyperbolas?
5. How do directrices relate to the properties of a hyperbola?

**Tip:** For hyperbolas, always check which variable is positive to determine the orientation (horizontal or vertical).

Describe and graph the following hyperbolas: (y + 1)^2/36 - (x - 1)^2/64 = 1 and 9x^2 - 4y^2 - 36 = 0.

Analyze and graph two hyperbola equations: (y + 1)^2/36 - (x - 1)^2/64 = 1 and 9x^2 - 4y^2 - 36 = 0. This solution includes key features such as center, vertices, co-vertices, foci, and asymptotes. Suitable for students in Grades 11-12.

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