https://file.aiquickdraw.com/mathbot/file/173067766055610ep5gx3.jpg

Let's go through each question to determine the answers.

### Question 1:
We are given:
- Vertices at \((0, 3)\) and \((0, -3)\).
- Asymptote equation: \( y = \pm \frac{1}{3} x \).

Since the vertices are aligned vertically, this hyperbola opens vertically and its standard form is:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1.
\]

1. **Distance between the vertices:** The distance between the vertices is 6, so \(2a = 6\), which gives \(a = 3\). Thus, \(a^2 = 9\).

2. **Finding \(b^2\):** The slope of the asymptotes is \(\pm \frac{a}{b} = \pm \frac{1}{3}\). Since \(a = 3\), we can solve for \(b\) as follows:
   \[
   \frac{3}{b} = \frac{1}{3} \Rightarrow b = 9 \Rightarrow b^2 = 81.
   \]

So, the equation of the hyperbola is:
\[
\frac{y^2}{9} - \frac{x^2}{81} = 1.
\]

**Answer for Question 1:** \(\frac{y^2}{9} - \frac{x^2}{81} = 1\).

---

### Question 2:
We are given:
- Center at the origin.
- Vertex at \((7, 0)\).
- Co-vertex at \((0, 3)\).

Since the vertex is along the x-axis, this hyperbola opens horizontally, and the standard form is:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.
\]

1. **Finding \(a^2\):** The distance from the center to the vertex is \(a = 7\), so \(a^2 = 49\).

2. **Finding \(b^2\):** The distance from the center to the co-vertex is \(b = 3\), so \(b^2 = 9\).

Therefore, the equation is:
\[
\frac{x^2}{49} - \frac{y^2}{9} = 1.
\]

**Answer for Question 2:** \(\frac{x^2}{49} - \frac{y^2}{9} = 1\).

---

### Question 3:
We need to find the value of \(a\) in the equation:
\[
\frac{y^2}{49} - \frac{x^2}{64} = 1.
\]

In this form, \(a^2 = 49\), so \(a = \sqrt{49} = 7\).

**Answer for Question 3:** \(7\).

---

### Question 4:
We are asked to identify the statement that is **not true** about the hyperbola:
\[
\frac{x^2}{4} - \frac{y^2}{16} = 1.
\]

For this hyperbola:
- The standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) indicates it opens horizontally.
- \(a^2 = 4\) (thus \(a = 2\)) and \(b^2 = 16\) (thus \(b = 4\)).
- **Vertices**: Since \(a = 2\), the vertices are at \((\pm 2, 0)\).
- **Co-vertices**: Since \(b = 4\), the co-vertices are at \((0, \pm 4)\).

Let's check each statement:
1. **Co-vertices** at \((0, 4)\) and \((0, -4)\) — Correct.
2. **Transverse axis** is horizontal — Correct (since it opens along the x-axis).
3. **Vertices** at \((2, 0)\) and \((-2, 0)\) — Correct.
4. **Asymptote slope \(4 : 2 = 2 : 1\)** — Incorrect. The correct slope is \(\pm \frac{b}{a} = \pm \frac{4}{2} = \pm 2\).

**Answer for Question 4:** "Asymptote slope \(4 : 2 = 2 : 1\)" (This statement is incorrect).

---

### Question 5:
We are asked for the **transverse axis** of the hyperbola:
\[
\frac{y^2}{16} - \frac{x^2}{36} = 1.
\]

In this form, \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), so the hyperbola opens **vertically**.

**Answer for Question 5:** Vertical.

---

Would you like any further details or have questions on any step? Here are five related questions to deepen understanding:

1. How do you determine if a hyperbola opens horizontally or vertically?
2. What is the relationship between the slopes of the asymptotes and the values of \(a\) and \(b\) in a hyperbola?
3. Why is it important to know the vertices and co-vertices when writing the equation of a hyperbola?
4. How can you verify the center of a hyperbola if given points are its vertices and co-vertices?
5. How do you find the focal points of a hyperbola from its standard equation?

**Tip:** Remember, for hyperbolas, the term with \(a^2\) corresponds to the axis along which the hyperbola opens, which helps in identifying its orientation.

The multiple questions on hyperbolas provided in the image, covering topics such as finding the standard form of a hyperbola, determining parameters like vertices, co-vertices, and the transverse axis, as well as understanding asymptotes.

Explore hyperbolas with detailed questions covering the standard form, asymptotes, and vertex calculations. This guide explains how to identify the orientation of a hyperbola, find vertices, co-vertices, and asymptote slopes. Perfect for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation