Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - 4.3 - Conics - 4.3 Exercises - Page 339: 74

Answer

Vertices: $(±12,0)$ Asymptotes: $y=±\frac{13}{12}x~~$ (Black lines)
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Work Step by Step

Standard form when transverse axis is horizontal: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $a^2=144$ $a=12$ $b^2=169$ $b=13$ Vertices when transverse axis is horizontal: $(±a,0)=(±12,0)$ Asymptotes when transverse axis is horizontal: $y=±\frac{b}{a}x$ $y=±\frac{13}{12}x$
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