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: 81b


$\frac{4\sqrt {61}}{13}~ft\approx2.4~ft$

Work Step by Step

A height of 18 feet is 5 feet above of the origin, that is, $y=5$ $x^2-\frac{3y^2}{169}=1$ $x^2-\frac{3(5)^2}{169}=1$ $x^2-\frac{75}{169}=1$ $x^2=\frac{75}{169}+1=\frac{244}{169}$ $x=±\frac{2\sqrt {61}}{13}$ Which gives us the points: $(±\frac{2\sqrt {61}}{13},5)$. The distance between these points is: $\frac{2\sqrt {61}}{13}-(-\frac{2\sqrt {61}}{13})=\frac{4\sqrt {61}}{13}\approx2.4$
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