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: 43

Answer

$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$

Work Step by Step

Here, $a=3$ and $c=5$ from the given points we can write $5^2=3^2+b^2$ $25-9=b^2$ $b^2=16$ The equation of the hyperbola is: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$
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