Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 26

Answer

$x=-\sqrt 2$

Work Step by Step

Consider $r =-\sqrt 2\sec \theta$ This can be re-arranged as: $r=\dfrac{-(\sqrt 2)}{\cos \theta}$ or, $r \cos \theta =-\sqrt 2 \implies x=-\sqrt 2$
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