Textbook Question
Graph each ellipse and give the location of its foci. (x +3)²/9 + (y -2)² = 1
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 47
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.
Verified step by step guidance1
Start with the given equation: \(4x^{2} - 9y^{2} - 16x + 54y - 101 = 0\).
Group the \(x\) terms and \(y\) terms together: \(4x^{2} - 16x - 9y^{2} + 54y = 101\).
Factor out the coefficients of the squared terms from each group: \(4(x^{2} - 4x) - 9(y^{2} - 6y) = 101\).
Complete the square for each group inside the parentheses:
- For \(x^{2} - 4x\), take half of \(-4\) which is \(-2\), square it to get \(4\), so add and subtract \(4\) inside the parentheses.
- For \(y^{2} - 6y\), take half of \(-6\) which is \(-3\), square it to get \(9\), so add and subtract \(9\) inside the parentheses.
Rewrite the equation including the completed squares and adjust the right side accordingly:
\(4(x^{2} - 4x + 4 - 4) - 9(y^{2} - 6y + 9 - 9) = 101\).
Then express the perfect square trinomials as squares:
\(4((x - 2)^{2} - 4) - 9((y - 3)^{2} - 9) = 101\).
Finally, distribute and move constants to the right side to isolate the squared terms.
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 terms. This technique helps convert the given equation into standard form, making it easier to identify the center and other properties of conic sections like hyperbolas.
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06:24
Solving Quadratic Equations by Completing the Square
Standard Form of a Hyperbola
The standard form of a hyperbola equation is either (x - h)²/a² - (y - k)²/b² = 1 or (y - k)²/a² - (x - h)²/b² = 1, where (h, k) is the center. Converting to this form reveals key features such as the orientation, vertices, and helps in graphing the hyperbola accurately.
Recommended video:
5:50Asymptotes of Hyperbolas
Foci and Asymptotes of a Hyperbola
The foci are two fixed points inside the hyperbola that define its shape, located using the relationship c² = a² + b². Asymptotes are lines that the hyperbola approaches but never touches, given by equations derived from the center and slopes ±b/a or ±a/b, guiding the graph's behavior at infinity.
Recommended video:
5:50Asymptotes of Hyperbolas
Related Practice
Textbook Question
Convert each equation to standard form by completing the square on x or y. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. y2 - 2y + 12x - 35 = 0
Textbook Question
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.
Textbook Question
Convert each equation to standard form by completing the square on x or y. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. x2 + 6x - 4y + 1 = 0
Textbook Question
Graph each ellipse and give the location of its foci. (x − 1)²/2 + (y +3)² /5= 1
Textbook Question
Identify each equation without completing the square. y2 - 4x + 2y + 21 = 0
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