Skip to main content
Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 7 - Conic SectionsProblem 49
Chapter 8, Problem 49

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. 4x2−25y2−32x+164=0 4x^2−25y^2−32x+164=0

Verified step by step guidance
1
Start with the given equation: \(4x^{2} - 25y^{2} - 32x + 164 = 0\).
Group the \(x\) terms and the \(y\) terms separately: \(4x^{2} - 32x - 25y^{2} + 164 = 0\).
Factor out the coefficient of \(x^{2}\) from the \(x\) terms: \(4(x^{2} - 8x) - 25y^{2} + 164 = 0\).
Complete the square for the expression inside the parentheses for \(x\): take half of \(-8\), which is \(-4\), square it to get \(16\), then add and subtract \(16\) inside the parentheses: \(4(x^{2} - 8x + 16 - 16) - 25y^{2} + 164 = 0\).
Rewrite the equation using the perfect square trinomial and simplify constants: \(4(x - 4)^{2} - 25y^{2} + 164 - 4 imes 16 = 0\), then move constants to the other side to isolate the conic.

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 a standard form, making it easier to identify the center, vertices, and other properties of conic sections like hyperbolas.
Recommended video:
06:24
Solving Quadratic Equations by Completing the Square

Standard Form of a Hyperbola

The standard form of a hyperbola is (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 the equation into this form reveals key features such as the center, vertices, and orientation, which are essential for graphing and further analysis.
Recommended video:
5:50
Asymptotes of Hyperbolas

Foci and Asymptotes of a Hyperbola

The foci are two fixed points inside the hyperbola that define its shape, located at a distance c from the center, where c² = a² + b². Asymptotes are lines that the hyperbola approaches but never touches, given by equations derived from the standard form. Identifying these helps in accurately sketching the hyperbola.
Recommended video:
5:50
Asymptotes of Hyperbolas
Related Practice
Textbook Question

Identify each equation without completing the square.

y2−4x−4y=0y^2-4x-4y=0

Textbook Question

Graph each ellipse and give the location of its foci. 9(x − 1)²+4(y+3)² = 36

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

3
views
Textbook Question

In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range. x29−y216=1\(\frac{x^2}{9}\) - \(\frac{y^2}{16}\) = 1