Explanation Given information: The parabola of an equation x 2 = 12 y and ( x − 3 ) 2 = 12 ( y − 1 ) . Definition used: A parabola with vertex at ( 0 , 0 ) and focus at ( 0 , p ) faces upwards if p > 0 and the focus is above the vertex or faces downwards if p < 0 and the focus is below the vertex. The directrix of the parabola is y = − p . The parabola equation in the standard form is, x 2 = 4 p y .............(1) A parabola with vertex at ( 0 , 0 ) and focus at ( p , 0 ) faces right if p > 0 and the focus is on the right side of the vertex or faces left if p < 0 and the focus is left side of the vertex. The directrix of the parabola is y = − p and The parabola equation in the standard form is, y 2 = 4 p x .............(2) A parabola with vertex at ( h , k ) and focus at ( h , k + p ) faces upwards if p > 0 or faces downwards if p < 0 . The parabola equation in the standard form is, ( x − h ) 2 = 4 p ( y − k ) .............(3). Here the directrix is y = − p + k and the parabola is vertically symmetric to the line x = h . A parabola with vertex at ( h , k ) and focus at ( h + p , k ) faces right if p > 0 or faces left if p < 0 . The parabola equation in the standard form is, ( y − k ) 2 = 4 p ( x − h ) .............(4). Here the directrix is x = − p + h and the parabola is horizontally symmetric to the line y = k . Calculation: The given equation of a parabola is x 2 = 12 y . Rewrite the equation of the parabola x 2 = 12 y as, x 2 = 4 p y x 2 = 12 y x 2 = 4 ( 3 ) y Compare the equation of parabola x 2 = 4 ( 3 ) y with the general equation of parabola x 2 = 4 p y to obtain the value of p. p = 3 Substitute 3 for p in the expression ( 0 , p ) to obtain the focus of the parabola as, focus = ( 0 , p ) = ( 0 , 3 ) Substitute 3 for p in the expression y = − p to obtain the directrix of the parabola as, y = − p y = − 3 . Substitute 0 for h and 0 for k in the expression ( h , k ) to obtain the vertex of the parabola as, vertex = ( h , k ) = ( 0 , 0 ) Therefore, the focus of the parabola is ( 0 , 3 ) , directrix y = − 3 and vertex is ( 0 , 0 ) is shown in figure 1

Bundle: College Algebra, 7th + WebAssign Printed Access Card for Stewart/Redlin/Watson's College Algebra, 7th Edition, Single-Term

7th Edition

ISBN: 9781305718944

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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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 7.4, Problem 2E

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The the focus, directrix and vertex of each parabola x2=12y and (x−3)2=12(y−1).

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Chapter 7.4, Problem 1E

Chapter 7.4, Problem 3E

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Chapter 7 Solutions

Bundle: College Algebra, 7th + WebAssign Printed Access Card for Stewart/Redlin/Watson's College Algebra, 7th Edition, Single-Term

Ch. 7.1 - A parabola is the set of all points in the plane...Ch. 7.1 - The graph of the equation x2=4py is a parabola...Ch. 7.1 - The graph of the equation y2=4px is a parabola...Ch. 7.1 - Label the focus, directrix, and vertex on the...Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Graphs of Parabolas Match the equation with the...Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The the focus, directrix and vertex of each parabola x 2 = 12 y and ( x − 3 ) 2 = 12 ( y − 1 ) .

Solution Summary: The author explains the focus, directrix, and vertex of each parabola of an equation x2=12y

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To find:

The the focus, directrix and vertex of each parabola x2=12y and (x−3)2=12(y−1).

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Chapter 7 Solutions

The the focus, directrix and vertex of each parabola x 2 = 12 y and ( x − 3 ) 2 = 12 ( y − 1 ) .

Solution Summary: The author explains the focus, directrix, and vertex of each parabola of an equation x2=12y

Concept explainers

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To find: The the focus, directrix and vertex of each parabola x2=12y and (x−3)2=12(y−1).

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Chapter 7 Solutions

To find:

The the focus, directrix and vertex of each parabola x2=12y and (x−3)2=12(y−1).