### Question 6Consider the surface in \( \mathbb{R}^3 \) defined by\[ x^2 + y^2 - 2z^2 + 6x - 8z = 7. \]#### (a) This surface is a quadric surface. Identify the surface by name. Briefly justify your answer.#### (b) Show that the point \( P(2, 1, 1) \) is on the surface.

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Answered: 6. Consider the surface in R³ defined…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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The general equation of a quadratic surface is given by: a₁x² + a₂y² + a₂z² + ²xy + ²x² + y² + ax + ay + a² + a₁0 = 0 10 Given nine points on this surface, it may be possible to determine its equation (a) Show that if the nine points (x,. y.) for i = 1, 2, 3,..., 9 lie on this surface, and if they determine uniquely the equation of this surface, then its equation can be written in determinant form as: z² xy z7 X1Y1 z3 z3 x y z x² x² x² x² an do an do co XZ XIZI ZX6Y6 X22 X2Z2 Y2Z2 X2 X3Y3 X3Z3 V3Z3 X3 X4Y4 X4Z4 Y4Z4 X4 Y4 x3y²z² Xsys XsZs YsZs Xs Ys X626 Y6Z6 X6 Y6 26 27 X7X7 X727 Y7Z7 X7 Y₁ Y8Z8 X8 Ys 1929 X9 Y9 zx8Y8 X8Z8 yz VIZI XI x y z X99 X929 1 1 Y2 22 1 3 Z3 1 Z4 1 Z5 1 1 Z7 1 Z8 1 Z9 1 N y Y₁ ZI = = 0 (b) Use the result in part (a) to determine the equation of the quadric surface that passes through the points (1, 2, 3), (2, 1, 7), (0. 4, 6). (3.-1, 4), (3, 0, 11), (-1, 5, 8), (9,-8, 3), (4. 5, 3), and (-2, 6, 10). (c) Classify the type of quadric surface in (b).
Let S be the quadric with equation 36x? –4y2 –9z² – 36x = 0. Sketch a graph of the surface with complete label of points on the graph. Follow the graph format of the given example. Example (0,0,8) (0,-4,0) -2,0,0) (2,0,0) (0,4,0)

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6. Consider the surface in R³ defined by x² + y² - 2z² + 6x - 8z = 7. (a) This surface is a quadric surface. Identify the surface by name. Briefly justify your answer. (b) Show that the point P(2, 1, 1) is on the surface.