## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 10 - Practice Exercises - Page 594: 59

#### Answer

$\dfrac{4}{1+2\cos \theta}$; equation of a hyperbola

#### Work Step by Step

The polar equation of a conic with eccentricity $e$ and directrix $x=k$ is: $r=\dfrac{ke}{1+e \cos \theta}$ ...(1) We are given that the vertices are: $e=2,k=2$ Then $x=2$ Thus, equation (1), becomes $r=\dfrac{4}{1+2\cos \theta}$; equation of a hyperbola

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