Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 10 - Parametric and Polar Curves - 10.4 Conic Sections - 10.4 Exercises - Page 751: 39

Answer

$$\eqalign{ & {\text{Vertices }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & {\text{Foci }}\left( { - \sqrt 5 ,0} \right){\text{ and }}\left( {\sqrt 5 ,0} \right) \cr & {\text{Asymptotes: }}y = \frac{1}{2}x{\text{ and }}y = - \frac{1}{2}x \cr} $$

Work Step by Step

$$\eqalign{ & \frac{{{x^2}}}{4} - {y^2} = 1 \cr & {\text{The equation is in the standard form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.{\text{ Then,}} \cr & \frac{{{x^2}}}{4} - {y^2} = 1,\,\,\,\,\,a = 2,\,\,\,b = 1 \cr & c = \sqrt {{a^2} + {b^2}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & \cr & {\text{With:}} \cr & {\text{Vertices }}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr & {\text{Vertices }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & {\text{Foci }}\left( { - c,0} \right){\text{ and }}\left( {c,0} \right) \cr & {\text{Foci }}\left( { - \sqrt 5 ,0} \right){\text{ and }}\left( {\sqrt 5 ,0} \right) \cr & {\text{Asymptotes: }}y = \frac{b}{a}x{\text{ and }}y = - \frac{b}{a}x \cr & {\text{Asymptotes: }}y = \frac{1}{2}x{\text{ and }}y = - \frac{1}{2}x \cr & \cr & {\text{Graph}} \cr} $$
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