   Chapter 10.6, Problem 5E

Chapter
Section
Textbook Problem

# Write a polar equation of a conic with the focus at the origin and the given data.5. Ellipse, eccentricity 2 3 , vertex (2, π)

To determine

To Find: The polar equation of a conic with the focus and origin.

Explanation

Given:

1. (i) Ellipse
2. (ii) Eccentricity 23
3. (iii) Vertex (2,π)

Calculation:

The polar equation for the given vertex is,

r=ed1±ecosθ

Case 1:

The curve appears horizontally which is left side due to vertex provided, so that the Directrix will be positive value.

The vertex lies at 2,π which is in horizontal axis.

Calculate the value of Directrix using provided vertex.

r=ed1ecosθ (1)

2=23d1(23cosπ)2=23d32cosπ32=23d×332cos(180)2=2d32cos(180)2=2d32(1)10=2dd=5

Substitute 5 for d in the equation (1).

r=ed1ecosθr=23×5123cosθr=10332cosθ3r=103×332cosθr=1032cosθ

Therefore, the polar equation of a conic with the eccentricity and vertex is 1032cosθ_

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