Chapter 10.6, Problem 17E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

# (a) Find the eccentricity and directrix of the conic r = 1/(1 − 2 sin θ) and graph the conic and its directrix. (b) If this conic is rotated counterclockwise about the origin through an angle 3π/4, write the resulting equation and graph its curve.

To determine

To Find: The eccentricity, equation of directrix and draw the conic and its directrix

Explanation

Given:

The given polar equation is r=112sinθ (1)

Calculation:

(a)

Calculate the eccentricity for the given equation.

The polar equation for the given equation will be either r=ed1±esinθ

Compare the given equation with the polar equation. Therefore,

e=2

Therefore, the eccentricity for the given equation is 2.

Here, the eccentricity is greater than one, so the conic is Hyperbola.

Write the equation of directrix.

From the equation (1),

ed=1 (2)

Substitute 2 for e in the equation (2).

ed=12×d=1d=12

Therefore, the equation of directrix is y=12

(b)

The value of x and y with respect to ‘r’ is,

x=rcosθy=rsinθ

Calculate the value of r for the various value of θ

r=112sinθ

Substitute 0° for θ in the above equation.

r=112sin(0°×π180)=1

Calculate the value of x.

x=rcosθ

Substitute 0° for θ and 1 for r in the above equation.

x=1×cos(0°×π180)=1

Calculate the value of y.

y=rsinθ

Substitute 0° for θ and 1 for r in the above equation

y=1×sin(0°×π180)=0

Similarly calculate the remaining values.

Tabulate the calculated value as shown in the table (1).

 θ r=11−2sinθ x=rcosθ y=rsinθ 0 1.00 1.00 0.00 10 1.53 1.51 0.27 20 3.16 2.97 1.08 40 -3.50 -2.68 -2.25 50 -1.88 -1.21 -1.44 60 -1.37 -0.68 -1.18 70 -1.14 -0.39 -1.07 80 -1.03 -0.18 -1.02 90 -1.00 0.00 -1.00 100 -1.03 0.18 -1.02 110 -1.14 0.39 -1.07 120 -1.37 0.68 -1.18 130 -1.88 1.21 -1.44 140 -3

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