   Chapter 10.6, Problem 54E

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.Hyperbola: focus at (5, 0); vertices at (4, 0), (4, π )

To determine

To calculate: The polar form of the equation of hyperbola having focus at (5,0) and vertices at (4,0),(4,π) using the result of exercise 50. That is the polar form of the equation for a hyperbola is r2=b21e2cos2θ.

Explanation

Given:

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

Formula used:

The rectangular form of the hyperbola is x2a2y2b2=1 where ab has foci (±c,0).

Here c2=a2+b2 eccentricity e=ca and vertices (±a,0) and its polar form is r2=b21e2cos2θ.

Calculation:

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

The rectangular form of the hyperbola is x2a2y2b2=1 here ab>0 has foci (±c,0)

Here c2=a2+b2 and vertices (±a,0)

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