Explanation Formula used: The hyperbola y 2 a 2 − x 2 b 2 = 1 has foci ( 0 , ± c ) , where c 2 = a 2 + b 2 , vertices ( 0 , ± a ) , and asymptotes y = ± ( a b ) x . Calculation: The given equation is, y 2 − 2 = x 2 − 2 x Rewrite the equation as follows. y 2 − x 2 + 2 x − 1 = 2 − 1 y 2 − ( x − 1 ) 2 = 1 y 2 1 − ( x − 1 ) 2 1 = 1 1 y 2 1 − ( x − 1 ) 2 1 = 1 This equation represents a hyperbola. Compute the center of the hyperbola using the equation. ( y − k ) 2 a 2 + ( x − h ) 2 b 2 = 1 ( y − 0 ) 2 1 + ( x − ( 1 ) ) 2 1 = 1 Therefore, the center of the hyperbola ( h , k ) is ( 1 , 0 )

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Author: Stewart

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Question

Chapter 10.5, Problem 30E

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To identify: The type of conic section from the equation y2−2=x2−2x and find out its vertices and foci.

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Chapter 10.5, Problem 29E

Chapter 10.5, Problem 31E

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Chapter 10 Solutions

Calculus, Early Transcendentals

Ch. 10.1 - Prob. 1E Ch. 10.1 - For the given parametric equations, find the...Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E

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The type of conic section from the equation y 2 − 2 = x 2 − 2 x and find out its vertices and foci.

Solution Summary: The author identifies the type of conic section from the equation y2-2=x 2x and finds out its vertices and foci.

To identify: The type of conic section from the equation y2−2=x2−2x and find out its vertices and foci.

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Chapter 10 Solutions