Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 10 - Analytic Geometry - 10.2 Ellipses - 10.2 Exercises - Page 978: 9

Answer

$$\eqalign{ & {\text{domain: }}\left[ { - 3,3} \right] \cr & {\text{range: }}\left[ { - 1,1} \right] \cr & {\text{center: }}\left( {0,0} \right) \cr & {\text{Vertices: }}\left( { \pm 3,0} \right) \cr & {\text{Endpoints of the minor axis:}}\left( {0, \pm 1} \right) \cr & {\text{Foci }}\left( { \pm 2\sqrt 2 ,0} \right) \cr} $$

Work Step by Step

$$\eqalign{ & \frac{{{x^2}}}{9} + {y^2} = 1 \cr & {\text{The equation of the ellipse is in the form }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,\left( {a > b} \right) \cr & a = 3,\,\,b = 1 \cr & c = \sqrt {{3^2} - {1^2}} = 2\sqrt 2 \cr & \cr & {\text{The ellipse with center at the origin}} \cr & {\text{Vertices }}\left( { \pm a,0} \right) \cr & {\text{Vertices }}\left( { \pm 3,0} \right) \cr & \cr & {\text{Foci }}\left( { \pm c,0} \right) \cr & {\text{Foci }}\left( { \pm 2\sqrt 2 ,0} \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ { - a,a} \right] \cr & {\text{domain }}\left[ { - 3,3} \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ { - b,b} \right] \cr & {\text{range }}\left[ { - 1,1} \right] \cr & \cr & {\text{Endpoints of the minor axis: }}\left( {0,\, \pm b} \right) = \left( {0, \pm 1} \right) \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 3,3} \right] \cr & {\text{range: }}\left[ { - 1,1} \right] \cr & {\text{center: }}\left( {0,0} \right) \cr & {\text{Vertices: }}\left( { \pm 3,0} \right) \cr & {\text{Endpoints of the minor axis:}}\left( {0, \pm 1} \right) \cr & {\text{Foci }}\left( { \pm 2\sqrt 2 ,0} \right) \cr} $$
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