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Let S be the quadric with equation 36x2 –4y² –9z² – 36x = 0. Describe the traces of S on the coordinate planes (follow the steps and format from the given example) Classify the graph of S and sketch a color-coded graph of the surface with complete label of points on the graph. %3|

Let S be the quadric with equation 36x2 –4y² –9z² – 36x = 0. Describe the traces of S on the coordinate planes (follow the steps and format from the given example) Classify the graph of S and sketch a color-coded graph of the surface with complete label of points on the graph. %3|

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter2: Analytic Trigonometry
Section2.3: Solving Trigonometric Equations
Problem 11ECP
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Please sketch the graph with all important labels and points.

Let S be the quadric with equation 36x? -4y? -9z² – 36x = 0. Describe the traces of S on the coordinate planes (follow the steps and format from the given example) • Classify the graph of S and sketch a color-coded graph of the surface with complete label of points on the graph. 2
Transcribed Image Text:Let S be the quadric with equation 36x? -4y? -9z² – 36x = 0. Describe the traces of S on the coordinate planes (follow the steps and format from the given example) • Classify the graph of S and sketch a color-coded graph of the surface with complete label of points on the graph. 2
Trace on the xy-plane Trace on the yz-plane Set z = 0 on the equation Set x = 0 on the equation x² y? 1 4 x² y? 1 4 |3D 8 16 16 We obtain We obtain x². y2 1. 16 y? 1- 16 8 4 Equivalently, This trace is an ellipse with the following properties: O Center at (0, 0,0). O Vertices are (0,0 ± v16, 0) = (0,±4,0). O The endpoints of the minor axis are (0± V4, 0,0) = (±2,0,0). y2 z = 8 – This trace is a parabola with the following properties: O Vertex at (0,0,8). O y-intercepts (set z = 0) are the solutions to %3D %3D y? 0 = 8 2 Trace on the xz-plane Graph of Elliptic Paraboloid Set y = 0 on the equation Finally, the graph above the xy-plane of the equation x2 1 4 y2 2z = 16 – 4x2 – y2 16 with its traces and important points: We obtain x2 :1- 4 ↑ ² 8. (0,0, 8) Equivalently, z = 8 – 2x2. This trace is a parabola with the following properties: O Vertex at (0,0,8). O x-intercepts (set z = 0) are the solutions to (0, –4, 0) (-2,0,0) 0 = 8 – 2x2. - That is, the points (±2,0,0). (2,0, 0) (0, 4, 0) || || || N 00
Transcribed Image Text:Trace on the xy-plane Trace on the yz-plane Set z = 0 on the equation Set x = 0 on the equation x² y? 1 4 x² y? 1 4 |3D 8 16 16 We obtain We obtain x². y2 1. 16 y? 1- 16 8 4 Equivalently, This trace is an ellipse with the following properties: O Center at (0, 0,0). O Vertices are (0,0 ± v16, 0) = (0,±4,0). O The endpoints of the minor axis are (0± V4, 0,0) = (±2,0,0). y2 z = 8 – This trace is a parabola with the following properties: O Vertex at (0,0,8). O y-intercepts (set z = 0) are the solutions to %3D %3D y? 0 = 8 2 Trace on the xz-plane Graph of Elliptic Paraboloid Set y = 0 on the equation Finally, the graph above the xy-plane of the equation x2 1 4 y2 2z = 16 – 4x2 – y2 16 with its traces and important points: We obtain x2 :1- 4 ↑ ² 8. (0,0, 8) Equivalently, z = 8 – 2x2. This trace is a parabola with the following properties: O Vertex at (0,0,8). O x-intercepts (set z = 0) are the solutions to (0, –4, 0) (-2,0,0) 0 = 8 – 2x2. - That is, the points (±2,0,0). (2,0, 0) (0, 4, 0) || || || N 00
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