College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Conic Sections - Exercise Set 7.2 - Page 687: 87



Work Step by Step

Yes, because if I plug in $y=-\frac{2}{3}x$ into the hyperbola's equation I get $\frac{x^2}{9}-\frac{(-\frac{2}{3}x)^2}{4}=\frac{x^2}{9}-\frac{\frac{4}{9}x^2}{4}=\frac{x^2}{9}-\frac{x^2}{9}=0\ne1$, thus they have no points in common
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