Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Conic Sections - 10.5 Exercises - Page 720: 42



Work Step by Step

We have $a^2-b^2=4^2=16$ or $a^2=16+b^2$ ... (1) Thus, $\dfrac{(-4)^{2}}{a^2}+\dfrac{(1.8)^{2}}{b^2}=1$ or, $16b^2+(3.24)a^2=a^2b^2$ or, $b^4-3.24b^2-51.84=0$ Thus, $b^2=9$ From equation (1), we have $a^2=16+b^2$ or, $a^2=16+9=25$ Hence, $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$
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