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: 38

Answer

$$\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{64}} = 1$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can see that the orientation of Major Axis is}} \cr & {\text{Horizontal }}\left( {{\text{along the x - axis}}} \right){\text{ and the ellipse is centered at the }} \cr & {\text{Origin}}.{\text{ Then}}{\text{, the equation is of the form}} \cr & \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr & \cr & {\text{The length of the major axis is 20}} \cr & {\text{2}}a = 20,\,\,\,\,a = 10 \cr & {\text{The length of the minor axis is 16}} \cr & 2b = 16,\,\,\,b = 8 \cr & \cr & \frac{{{x^2}}}{{{{\left( {10} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( 8 \right)}^2}}} = 1 \cr & \frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{64}} = 1 \cr} $$
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