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: 27

Answer

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Work Step by Step

We can write this equation in standard form as $\displaystyle \frac{x^{2}}{1}-\frac{y^{2}}{1}=1$ When in this form, the hyperbola has foci on the x-axis: $\quad \displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\quad a=1, \ b=1$ Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{1+1}=\sqrt{2}$ Foci: $\quad (\pm c, 0)=\quad (\pm\sqrt{2}, 0)$ Vertices: $\quad (\pm a, 0)=\quad (\pm 1 0)$ Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x=\pm\frac{1}{1}x$ $y=\pm x$
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