Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.6 - Conic Sections - Exercises 11.6 - Page 678: 22

Answer

See image:

Work Step by Step

Divide the equation with $90$ $\displaystyle \frac{x^{2}}{10}+\frac{y^{2}}{9}=1$ $\displaystyle \frac{x^{2}}{(\sqrt{10})^{2}}+\frac{y^{2}}{(3)^{2}}=1$, which is of the form $\quad \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad (a\gt b)$, $\Rightarrow$ the foci are on the x-axis. Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}=\sqrt{10-9}=1$ Foci: $\quad(\pm c, 0)= \quad(\pm 1, 0)$ Vertices: $\quad (\pm a, 0)= \quad (\pm\sqrt{10}, 0)$
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