(a) Explanation Definition used: Definition 1: “The equation of the shifted hyperbola with center as ( h , k ) and horizontal axis (x-axis) axis as transverse axis is ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 , when a > 0 , b > 0 , c 2 = a 2 + b 2 ”. Definition 2: “The equation of the hyperbola with center at the origin, vertices ( 0 , − a ) , ( 0 , a ) and foci F 1 ( 0 , − c ) , F 2 ( 0 , c ) is y 2 a 2 − x 2 b 2 = 1 , where c = a 2 + b 2 , a > 0 , b > 0 ”. Definition 3: “The equation of the asymptotes of the hyperbola ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 are ( y − k ) a + ( x − h ) b = 0 and ( y − k ) a − ( x − h ) b = 0 ”. Definition 4: “If h and k are any two positive real numbers and a graph of any equation is in terms of x and y then. ( i ) Replacing x by x − h , shift the graph right to h units. ( i i ) Replacing x by x + h , shift the graph left to h units. ( i i i ) Replacing y by y − k , shift the graph upward to k units. ( i v ) Replacing y by y + k , shift the graph downward to k units”. The given equation of conic is, y 2 = x 2 + 6 y . It can be written as follows. y 2 = x 2 + 6 y y 2 − 6 y = x 2 y 2 − 6 y + 9 − 9 − x 2 = 0 ( y − 3 ) 2 − x 2 = 9 Thus, the equation is ( y − 3 ) 2 9 − x 2 9 = 1 (1) Now, by the definition 1, the equation (1) is a shifted hyperbola of y 2 9 − x 2 9 = 1 (2) That is, equation ( 1 ) may obtain from equation ( 2 ) replacing x by x and y by y − 3 . Therefore, first find the center, vertices and foci of the hyperbola ( 2 ) and keep the x -coordinates as same, and y -coordinates by 3 unit upward. Compare the equation (2) with the hyperbola y 2 a 2 − x 2 b 2 = 1 (b) To determine To sketch: The graph of the hyperbola y 2 = x 2 + 6 y .

Precalculus: Mathematics for Calculus (Standalone Book)

7th Edition

ISBN: 9781305071759

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Brooks Cole

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Chapter 11, Problem 36RE

(a)

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola y2=x2+6y .

(b)

To determine

To sketch: The graph of the hyperbola y2=x2+6y .

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Chapter 11 Solutions

Precalculus: Mathematics for Calculus (Standalone Book)

Ch. 11.1 - A parabola is the set of all points in the plane...Ch. 11.1 - The graph of the equation x2 = 4py is a parabola...Ch. 11.1 - The graph of the equation y2 = 4px is a parabola...Ch. 11.1 - Label the focus, directrix, and vertex on the...Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Prob. 6E Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Prob. 8E Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Graphs of Parabolas Match the equation with the...

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The center, vertices, foci and asymptotes of the hyperbola y 2 = x 2 + 6 y .

Solution Summary: The author explains the equation of the shifted hyperbola with center at the origin, vertices, foci, and asymptotes.

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To find: The center, vertices, foci and asymptotes of the hyperbola y2=x2+6y .

To sketch: The graph of the hyperbola y2=x2+6y .

Chapter 11 Solutions