Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 18


$$ \left(\frac{y}{6}\right)^{2}-\left(\frac{x}{12\sqrt 2}\right)^{2}=1. $$

Work Step by Step

Since the vertices are $(0,\pm 6)$ and $e=3$, then the axis of the hyperbola is the y-axis and then $b=6$, moreover, $ e=\frac{c}{b}=\frac{c}{6}=3 $, and hence $c=18 $ and $a=\sqrt{c^2-b^2}=\sqrt{18^2-6^2}=12\sqrt{2}$. Hence the equation of the hyperbola is given by $$ \left(\frac{y}{6}\right)^{2}-\left(\frac{x}{12\sqrt 2}\right)^{2}=1. $$
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