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: 42

Answer

$$\eqalign{ & {\text{Vertices }}\left( {0, - 2} \right){\text{ and }}\left( {0,2} \right) \cr & {\text{Foci }}\left( {0, - \sqrt {29} } \right){\text{ and }}\left( {0,\sqrt {29} } \right) \cr & {\text{Asymptotes: }}y = \frac{2}{5}x{\text{ and }}y = - \frac{2}{5}x \cr} $$

Work Step by Step

$$\eqalign{ & 25{y^2} - 4{x^2} = 100 \cr & {\text{Divide the equation by 100}} \cr & \frac{{{y^2}}}{4} - \frac{{{x^2}}}{{25}} = 1 \cr & {\text{The equation is in the standard form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1.{\text{ Then,}} \cr & \frac{{{y^2}}}{4} - \frac{{{x^2}}}{{25}} = 1,\,\,\,\,\,a = 2,\,\,\,b = 5 \cr & c = \sqrt {{a^2} + {b^2}} = \sqrt {4 + 25} = \sqrt {29} \cr & \cr & {\text{With:}} \cr & {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr & {\text{Vertices }}\left( {0, - 2} \right){\text{ and }}\left( {0,2} \right) \cr & {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr & {\text{Foci }}\left( {0, - \sqrt {29} } \right){\text{ and }}\left( {0,\sqrt {29} } \right) \cr & {\text{Asymptotes: }}y = \frac{a}{b}x{\text{ and }}y = - \frac{a}{b}x \cr & {\text{Asymptotes: }}y = \frac{2}{5}x{\text{ and }}y = - \frac{2}{5}x \cr & \cr & {\text{Graph}} \cr} $$
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