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: 37

Answer

$$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can see that the orientation of Major Axis is}} \cr & {\text{vertical }}\left( {{\text{along the y - axis}}} \right){\text{ and the ellipse is centered at the }} \cr & {\text{Origin}}.{\text{ Then, the equation is of the form}} \cr & \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr & \cr & {\text{The length of the major axis is 6}} \cr & {\text{2}}a = 6,\,\,\,\,a = 3 \cr & {\text{The length of the minor axis is 4}} \cr & 2b = 4,\,\,\,b = 2 \cr & \cr & \frac{{{x^2}}}{{{{\left( 2 \right)}^2}}} + \frac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \cr & \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1 \cr} $$
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