Determine the equations of the asymptotes of the following hyperbola. You will need to rewrite the equation in standard form first. 9x² - 4y² + 18x - 24y - 63 = 0
Determine the equations of the asymptotes of the following hyperbola. You will need to rewrite the equation in standard form first. 9x² - 4y² + 18x - 24y - 63 = 0
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Title: Finding Asymptotes of a Hyperbola**
**Introduction:**
In this lesson, we will determine the equations of the asymptotes of a given hyperbola. To do this, we will first need to rewrite the hyperbola's equation in its standard form.
**Problem Statement:**
Determine the equations of the asymptotes of the following hyperbola. You need to rewrite the equation in standard form first.
\[ 9x^2 - 4y^2 + 18x - 24y - 63 = 0 \]
**Steps to Solve:**
1. **Rewrite the Equation in Standard Form:**
- Begin by organizing the given equation by grouping like terms:
\[ 9x^2 + 18x - 4y^2 - 24y = 63 \]
2. **Complete the Square:**
- In order to rewrite the equation in standard form, complete the square for the \(x\)-terms and the \(y\)-terms.
For the \(x\)-terms: \(9x^2 + 18x\)
\[ 9(x^2 + 2x) = 9(x^2 + 2x + 1 - 1) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9 \]
For the \(y\)-terms: \(-4y^2 - 24y\)
\[ -4(y^2 + 6y) = -4(y^2 + 6y + 9 - 9) = -4((y+3)^2 - 9) = -4(y+3)^2 + 36 \]
3. **Substitute Back into the Equation:**
\[ 9(x+1)^2 - 9 - 4(y+3)^2 + 36 = 63 \]
Simplify:
\[ 9(x+1)^2 - 4(y+3)^2 + 27 = 63 \]
\[ 9(x+1)^2 - 4(y+3)^2 = 36 \]
4. **Divide by 36 to Standardize:**
\[ \frac{9(x+1)^2}{36} - \frac{4(y+3)^2}{36} = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F585970c1-8db0-450f-987b-10f4de58f48c%2F2fc8d22b-d625-43d8-afa2-6701103fc118%2Fh9e6m3q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Finding Asymptotes of a Hyperbola**
**Introduction:**
In this lesson, we will determine the equations of the asymptotes of a given hyperbola. To do this, we will first need to rewrite the hyperbola's equation in its standard form.
**Problem Statement:**
Determine the equations of the asymptotes of the following hyperbola. You need to rewrite the equation in standard form first.
\[ 9x^2 - 4y^2 + 18x - 24y - 63 = 0 \]
**Steps to Solve:**
1. **Rewrite the Equation in Standard Form:**
- Begin by organizing the given equation by grouping like terms:
\[ 9x^2 + 18x - 4y^2 - 24y = 63 \]
2. **Complete the Square:**
- In order to rewrite the equation in standard form, complete the square for the \(x\)-terms and the \(y\)-terms.
For the \(x\)-terms: \(9x^2 + 18x\)
\[ 9(x^2 + 2x) = 9(x^2 + 2x + 1 - 1) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9 \]
For the \(y\)-terms: \(-4y^2 - 24y\)
\[ -4(y^2 + 6y) = -4(y^2 + 6y + 9 - 9) = -4((y+3)^2 - 9) = -4(y+3)^2 + 36 \]
3. **Substitute Back into the Equation:**
\[ 9(x+1)^2 - 9 - 4(y+3)^2 + 36 = 63 \]
Simplify:
\[ 9(x+1)^2 - 4(y+3)^2 + 27 = 63 \]
\[ 9(x+1)^2 - 4(y+3)^2 = 36 \]
4. **Divide by 36 to Standardize:**
\[ \frac{9(x+1)^2}{36} - \frac{4(y+3)^2}{36} = 1
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