Domanda del libro di testo
Graph each ellipse and give the location of its foci. (x +3)²/9 + (y -2)² = 1
794
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Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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TOC.tableOfContents
- 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
Capitolo 8, Problema 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.
Guida verificata passo dopo passo1
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.
Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
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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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.
Video consigliato:
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.
Video consigliato:
5:50Asymptotes of Hyperbolas
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