   Chapter 12.6, Problem 37E

Chapter
Section
Textbook Problem

# Reduce the equation to one of the standard forms, classify the surface, and sketch it.x2 − y2 + z2 − 4x − 2z = 0

To determine

To classify: The given surface equation and the surface equation.

Explanation

Given data:

Surface equation is x2y2+z24x2z=0 .

The surface equation x2y2+z24x2z=0 is sketched.

Formula used:

Consider the standard equation of hyperboloid of one sheet along the x axis.

x2a2y2b2+z2c2=1 (1)

Consider the given surface equation.

x2y2+z24x2z=0 (2)

Rearrange the equation.

(x24x+4)y2+(z22z+1)=0+4+1(x2)2y2+(z1)2=5(x2)2y2+(z1)25=1

(x2)25y25+(z1)25=1 (3)

Find the center by using equation (3).

(x2)2=0x2=0x=2

y2=0y=0

(z1)2=0z1=0z=1

As the computed expression (3) satisfies the equation of hyperboloid of one sheet, which is centered at (2,0,1) .

Thus, the surface equation x2y2+z24x2z=0 is a hyperboloid of one sheet.

Find a, b, and c by comparing equation (3) with equation (1).

a2=5a=5a=±2.23

The value of a is ±2.23 .

b2=5b=5b=±2.23

The value of b is ±2.23 .

c2=5c=5c=±2.23

The value of c is ±2.23 .

Case i:

Let x=k .

Substitute k for x in equation (3),

(k2)25y25+(z1)25=1

y25+(z1)25=1(k2)25 (4)

Modify equation (4) for k=4.25 ,

y25+(z1)25=1(4.252)25y25+(z1)25=1(2.25)25y25+(z1)25=155y25+(z1)25=11

Simplify the equation.

y25+(z1)25=0

Substitute 1 for z,

y25+(11)25=0y25=0y2=0y=0

So the hyperboloid lies at (y,z)=(0,1) .

Hence, the surface equation has a point (y,z)=(0,1) for k=4.25 .

Substitute 0 for y and 1 for z in equation (4),

025+(11)25=1(k2)250+0=1(k2)25(k2)25=1(k2)2=5

Modify the equation,

(k2)=5k=±2

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