Explanation Equation used: “ Hyperbola: The general form of the equation of the hyperbola in xy -coordinate plane is, ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 ( transverse axis parallel to x -axis ) ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 ( transverse axis parallel to y -axis ) where ( h , k ) is the center.” Calculation: The given equation of the curve is, 5 x 2 + 12 y + 18 = 2 y 2 Identify the type of the curve from the given equation as follows. 5 x 2 + 12 y + 18 = 2 y 2 5 x 2 + 12 y + 18 − 2 y 2 = 0 2 ( y 2 − 6 y + 9 ) − 5 x 2 = 18 + 18 ( y − 3 ) 2 18 − 5 x 2 36 = 1 It is observed from the obtained curve that, it is a hyperbola with transverse axis parallel to the y- axis

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Chapter 21.7, Problem 38E

To determine

The center or vertex (if it is a parabola) of the given curve and to sketch the curve.

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Chapter 21.7, Problem 37E

Chapter 21.7, Problem 39E

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Chapter 21 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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