Textbook Question
In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range.
Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 8, Problem 53
Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 9x² + 16y² – 18x + 64y – 71 = 0
Verified step by step guidance1
Start with the given equation: \(9x^{2} + 16y^{2} - 18x + 64y - 71 = 0\).
Group the \(x\) terms and \(y\) terms together: \(9x^{2} - 18x + 16y^{2} + 64y = 71\).
Factor out the coefficients of the squared terms from each group: \(9(x^{2} - 2x) + 16(y^{2} + 4y) = 71\).
Complete the square for each group inside the parentheses:
- For \(x^{2} - 2x\), take half of \(-2\) which is \(-1\), square it to get \(1\), so add and subtract \(1\) inside the parentheses.
- For \(y^{2} + 4y\), take half of \(4\) which is \(2\), square it to get \(4\), so add and subtract \(4\) inside the parentheses.
Rewrite the equation including the completed squares and adjust the right side accordingly:
\(9(x^{2} - 2x + 1 - 1) + 16(y^{2} + 4y + 4 - 4) = 71\).
Then express the perfect square trinomials as squares:
\(9((x - 1)^{2} - 1) + 16((y + 2)^{2} - 4) = 71\).
Finally, distribute and move constants to the right side to isolate the squared terms.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Completing the Square
Completing the square is a method used to rewrite quadratic expressions in the form (x - h)² or (y - k)² by adding and subtracting appropriate constants. This technique helps transform the given equation into a standard form, making it easier to identify the center and dimensions of conic sections like ellipses.
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Standard Form of an Ellipse
The standard form of an ellipse equation is (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center, and a and b are the lengths of the semi-major and semi-minor axes. Converting to this form allows for straightforward graphing and analysis of the ellipse's properties.
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Foci of an Ellipse
The foci are two fixed points inside an ellipse such that the sum of the distances from any point on the ellipse to the foci is constant. Their locations depend on the values of a, b, and the center, and are found using the relationship c² = |a² - b²|, where c is the distance from the center to each focus.
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Related Practice
Textbook Question
Identify each equation without completing the square.
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Textbook Question
In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range.