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 750: 32

Answer

$$\eqalign{ & {\text{Vertices: }}\left( {0, \pm 2\sqrt 3 } \right);\,\,{\text{Foci:}}\,\,\left( {0, \pm \sqrt 5 } \right) \cr & {\text{major axis has length }}4\sqrt 3 \cr & {\text{minor axis has length }}2\sqrt 5 \cr} $$

Work Step by Step

$$\eqalign{ & 12{x^2} + 5{y^2} = 60 \cr & {\text{Divide both sides by 60}} \cr & \frac{{12{x^2}}}{{60}} + \frac{{5{y^2}}}{{60}} = \frac{{60}}{{60}} \cr & \frac{{{x^2}}}{5} + \frac{{{y^2}}}{{12}} = 1 \cr & \cr & {\text{The equation has the standard form}} \cr & \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1,\,\,\,\,\,a > b > 0 \cr & \frac{{{x^2}}}{5} + \frac{{{y^2}}}{{12}} = 1 \Rightarrow a = 2\sqrt 3 ,b = \sqrt 5 \cr & c = \sqrt {{a^2} - {b^2}} = \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}} = \sqrt 7 \cr & {\text{With}} \cr & {\text{Orientation of Major Axis Horizontal along the }}x{\text{ - axis, then}} \cr & {\text{Vertices }}\left( {0, - a} \right){\text{ and }}\left( {0,a} \right) \cr & {\text{Vertices }}\left( {0, - 2\sqrt 3 } \right){\text{ and }}\left( {0,2\sqrt 3 } \right) \cr & \cr & {\text{Foci }}\left( {0, - c} \right){\text{ and }}\left( {0,c} \right) \cr & {\text{Foci}}\left( {0, - \sqrt 5 } \right){\text{ and }}\left( {0,\sqrt 5 } \right) \cr & \cr & {\text{Length of the minor axis}} \cr & 2b = 2\left( {\sqrt 5 } \right) = 2\sqrt 5 \cr & {\text{Length of the major axis}} \cr & 2a = 2\left( {2\sqrt 3 } \right) = 4\sqrt 3 \cr & \cr & {\text{Graph }} \cr} $$
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