find the standard equation and graph 9x²-4y²-36x-24y-36=0 Determine the center, the lines which contains the transverse and conjugate axis,vertices,the foci and the equation of the asymptotes

To solve the problem, we need to identify the standard form of the equation and deduce relevant characteristics for the hyperbola. Here's the step-by-step process:

### Given Equation:
\[
9x^2 - 4y^2 - 36x - 24y - 36 = 0
\]

### Step 1: Rearrange the terms
Group the \(x\) and \(y\) terms together:
\[
9(x^2 - 4x) - 4(y^2 + 6y) = 36
\]

### Step 2: Complete the square
We'll complete the square for both the \(x\) and \(y\) terms inside the parentheses.

For \(x^2 - 4x\):
1. Take half of \(-4\), square it: \(\left(\frac{-4}{2}\right)^2 = 4\).
2. Add and subtract 4 inside the parentheses.

For \(y^2 + 6y\):
1. Take half of 6, square it: \(\left(\frac{6}{2}\right)^2 = 9\).
2. Add and subtract 9 inside the parentheses.

Thus, we get:
\[
9[(x^2 - 4x + 4) - 4] - 4[(y^2 + 6y + 9) - 9] = 36
\]
\[
9[(x - 2)^2 - 4] - 4[(y + 3)^2 - 9] = 36
\]
Distribute the 9 and -4:
\[
9(x - 2)^2 - 36 - 4(y + 3)^2 + 36 = 36
\]
Simplify:
\[
9(x - 2)^2 - 4(y + 3)^2 = 36
\]

### Step 3: Convert to standard form
Divide the entire equation by 36:
\[
\frac{9(x - 2)^2}{36} - \frac{4(y + 3)^2}{36} = 1
\]
Simplify:
\[
\frac{(x - 2)^2}{4} - \frac{(y + 3)^2}{9} = 1
\]

This is the standard form of a hyperbola:
\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]
where \( (h, k) \) is the center, \( a^2 \) is the square of the semi-major axis, and \( b^2 \) is the square of the semi-minor axis.

### Step 4: Identify the components
- **Center**: \((h, k) = (2, -3)\)
- **Transverse axis**: Horizontal because the \(x\)-term is positive
- **Vertices**: The distance from the center to the vertices is \(a = 2\). So the vertices are:
  - \( (2 + 2, -3) = (4, -3) \)
  - \( (2 - 2, -3) = (0, -3) \)
- **Foci**: The distance from the center to the foci is given by \(c\), where \(c^2 = a^2 + b^2 = 4 + 9 = 13\), so \(c = \sqrt{13}\). The foci are:
  - \( (2 + \sqrt{13}, -3) \)
  - \( (2 - \sqrt{13}, -3) \)
- **Asymptotes**: The equations of the asymptotes for a hyperbola are:
  - \( y - k = \pm \frac{b}{a} (x - h) \)
  - Here, \( b = 3 \), \( a = 2 \), \( h = 2 \), and \( k = -3 \). So, the equations of the asymptotes are:
    \[
    y + 3 = \pm \frac{3}{2} (x - 2)
    \]

### Final Answer Summary:
- **Standard Equation**: \(\frac{(x - 2)^2}{4} - \frac{(y + 3)^2}{9} = 1\)
- **Center**: \((2, -3)\)
- **Vertices**: \((4, -3)\) and \((0, -3)\)
- **Foci**: \((2 + \sqrt{13}, -3)\) and \((2 - \sqrt{13}, -3)\)
- **Asymptotes**: \(y + 3 = \frac{3}{2}(x - 2)\) and \(y + 3 = -\frac{3}{2}(x - 2)\)

Would you like a graph of this hyperbola?

### Related Questions:
1. How do we determine the transverse and conjugate axes for a hyperbola?
2. How is the distance between the foci calculated for any hyperbola?
3. How do the equations of the asymptotes change for vertical transverse axes?
4. What is the significance of completing the square in conic sections?
5. Can you explain how to find the eccentricity of a hyperbola?

**Tip**: The distance between the foci (denoted by \(2c\)) helps quantify the "stretch" of a hyperbola.

Explore the standard equation and graph of a hyperbola 9x² - 4y² - 36x - 24y - 36 = 0. Understand the center, lines containing the transverse and conjugate axes, vertices, foci, and equations of asymptotes with a step-by-step solution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation