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 635: 4

Answer

The vertices are $(\pm12,0)$. The foci are $(\pm \sqrt{468},0)$.

Work Step by Step

The equation $$ \left(\frac{x}{2}\right)^{2}-\left(\frac{y}{3}\right)^{2}=36\Longrightarrow\left(\frac{x}{12}\right)^{2}-\left(\frac{y}{18}\right)^{2}=1 $$ is hyperbola. We have $a=12$, $b=18$, $c=\sqrt{a^2+b^2}=\sqrt{468}$ and hence: The vertices are $(\pm12,0)$. The foci are $(\pm \sqrt{468},0)$.
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