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: 42



Work Step by Step

$a^2-b^2=4^2$ $a^2-b^2=16$ $\frac{(-4)^{2}}{a^2}+\frac{(1.8)^{2}}{b^2}=1$ $16b^2+3.24a^2=a^2b^2$ use $a^2-b^2=16$ Thus, $b^4-3.24b^2-51.84=0$ $b=3$ Now, $a^2=16+b^2$ $a=5$ Thus, $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
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