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The vertices of a hyperbola are at (-5, -2) and (-5, 12), and the point (-5, 30) is one of its foci. What is the equation of this hyperbo O A. (y – 5)? (= + 5)? -=D1 72 242 O B. ( + 5)2 (y – 5)2 242 72 -=1 Oc ( + 5)2 (3y – 5)2 72 =1 242 O D. (y- 5)2 ( + 5)? _', =D1 72 242

The vertices of a hyperbola are at (-5, -2) and (-5, 12), and the point (-5, 30) is one of its foci. What is the equation of this hyperbo O A. (y – 5)? (= + 5)? -=D1 72 242 O B. ( + 5)2 (y – 5)2 242 72 -=1 Oc ( + 5)2 (3y – 5)2 72 =1 242 O D. (y- 5)2 ( + 5)? _', =D1 72 242

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The vertices of a hyperbola are at (-5, -2) and (-5, 12), and the point (-5, 30) is one of its foci. What is the equation of this hyperbola? O A. (y – 5)2 (± + 5)² -=1 72 242 (= + 5)2_(y – 5)? OB. 242 72 =1 (= + 5)? (y – 5)² =1 C. 72 242 (y - 5)2 ( + 5)2 =1 OD. 72 242
Transcribed Image Text:The vertices of a hyperbola are at (-5, -2) and (-5, 12), and the point (-5, 30) is one of its foci. What is the equation of this hyperbola? O A. (y – 5)2 (± + 5)² -=1 72 242 (= + 5)2_(y – 5)? OB. 242 72 =1 (= + 5)? (y – 5)² =1 C. 72 242 (y - 5)2 ( + 5)2 =1 OD. 72 242
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