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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)
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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Question
**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
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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