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. Find the vertices of the hyperbola. 16x2 - 9y? = 144 %3D (0, O (0, - 3) and (0, 3) O (0, - 4) and (0, 4) O (-4, 0) and (4, 0) O (-3, 0) and (3, 0)

. Find the vertices of the hyperbola. 16x2 - 9y? = 144 %3D (0, O (0, - 3) and (0, 3) O (0, - 4) and (0, 4) O (-4, 0) and (4, 0) O (-3, 0) and (3, 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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**Problem 10: Find the vertices of the hyperbola.** Given the hyperbola equation: \[ 16x^2 - 9y^2 = 144 \] Select the correct pair of vertices from the options below: a) \((0, -3)\) and \((0, 3)\) b) \((0, -4)\) and \((0, 4)\) c) \((-4, 0)\) and \((4, 0)\) d) \((-3, 0)\) and \((3, 0)\) **Explanation:** To find the vertices, we start by rewriting the equation in its standard form. Divide the entire equation by 144: \[ \frac{16x^2}{144} - \frac{9y^2}{144} = 1 \] Simplify the fractions: \[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \] This is the standard form of a hyperbola centered at the origin, with a horizontal transverse axis. The vertices are determined by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 9\). Therefore, \(a = 3\). Thus, the vertices are at \((-3, 0)\) and \((3, 0)\). The correct answer is option d).
Transcribed Image Text:**Problem 10: Find the vertices of the hyperbola.** Given the hyperbola equation: \[ 16x^2 - 9y^2 = 144 \] Select the correct pair of vertices from the options below: a) \((0, -3)\) and \((0, 3)\) b) \((0, -4)\) and \((0, 4)\) c) \((-4, 0)\) and \((4, 0)\) d) \((-3, 0)\) and \((3, 0)\) **Explanation:** To find the vertices, we start by rewriting the equation in its standard form. Divide the entire equation by 144: \[ \frac{16x^2}{144} - \frac{9y^2}{144} = 1 \] Simplify the fractions: \[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \] This is the standard form of a hyperbola centered at the origin, with a horizontal transverse axis. The vertices are determined by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 9\). Therefore, \(a = 3\). Thus, the vertices are at \((-3, 0)\) and \((3, 0)\). The correct answer is option d).
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