Precalculus (6th Edition)

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

Chapter 10 - Analytic Geometry - 10.4 Summary of the Conic Sections - 10.4 Exercises - Page 996: 14

Answer

$${\text{ellipse}}$$

Work Step by Step

$$\eqalign{ & \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{36}} = 1 \cr & {\text{or}} \cr & \frac{{{x^2}}}{{{{\left( 5 \right)}^2}}} + \frac{{{y^2}}}{{{{\left( 6 \right)}^2}}} = 1 \cr & {\text{The equation is written in the form }}\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\,\,\,\,\left( {{\text{Where }}a > b} \right) \cr & {\text{Therefore,}} \cr & {\text{The graph of the equation }}\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{36}} = 1{\text{ is an ellipse}} \cr} $$
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