Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 48

Answer

$\dfrac{5(y-4)^{2}}{16}-\dfrac{5(x-2)^{2}}{64}=1$

Work Step by Step

Asymptotes are $y=3+1/2x, y=5-1/2x$ which gives $\frac{a}{b}=\frac{1}{2}$ $2a=b$ $a^2+(2a)^2=4^2$ $a=\frac{4}{\sqrt 5}$ and $b=\frac{8}{\sqrt 5}$ The equation of the hyperbola is: $\dfrac{5(y-4)^{2}}{16}-\dfrac{5(x-2)^{2}}{64}=1$
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