   Chapter 10.6, Problem 53E

Chapter
Section
Textbook Problem

# Finding a Polar Equation In Exercises 51-54, use the results of Exercises 49 and 50 to write the polar form of the equation of the conic.Ellipse: focus at (4, 0); vertices at (5, 0), (5, π )

To determine

To calculate: The polar form of the equation of ellipse having focus at (4,0) and vertices at (5,0),(5,π) using the result of question 49. That is the polar form of the equation for an ellipse is: r2=b21e2cos2θ

Explanation

Given:

The conic is ellipse having the focus at (4,0) and the vertices at (5,0),(5,π).

Formula used:

The rectangular form of the ellipse is x2a2+y2b2=1 where ab has foci (±c,0).

Here c2=a2b2, eccentricity e=ca and vertices (±a,0) and its polar form is r2=b21e2cos2θ.

Calculation:

The focus of the ellipse is at (4,0) and vertices are at (5,0),(5,π).

While the rectangular form of the ellipse is x2a2+y2b2=1 here ab>0 has foci at (±c,0)

Here c2=a2b2 and vertices (±a,0)

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