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 677: 5

Answer

$\displaystyle \frac{x^{2}}{4}-\frac{\mathrm{y}^{2}}{9}=1,\ \ \ $ (hyperbola) Foci: $\qquad (\pm\sqrt{13}, 0)$ Asymptotes:$\displaystyle \quad y=\pm\frac{3}{2}x$

Work Step by Step

Hyperbola, horizontal axis. Foci on the x-axis: $\quad \displaystyle \frac{x^{2}}{a^{2}}-\frac{\mathrm{y}^{2}}{b^{2}}=1$ The equation of this form is $\quad \displaystyle \frac{x^{2}}{4}-\frac{\mathrm{y}^{2}}{9}=1$ We read: $a=2,\quad b=3$ Foci: $(\pm c, 0),\qquad c=\sqrt{a^{2}+b^{2}}$ $c=\sqrt{4+13}=\sqrt{13}$ So the foci are at $(\pm\sqrt{13}, 0)$ Asymptotes:$\displaystyle \quad y=\pm\frac{b}{a}x$ $y=\displaystyle \pm\frac{3}{2}x$
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