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: 35

Answer

$${\text{hyperbola}}$$

Work Step by Step

$$\eqalign{ & {x^2} = 25 + {y^2} \cr & {\text{Subtract }}{y^2} \cr & {x^2} - {y^2} = 25 \cr & {\text{Divide by 25}} \cr & \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{25}} = 1 \cr & {\text{The equation is in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr & {\text{Therefore, the equation represents a hyperbola}} \cr & {\text{Graph}} \cr} $$
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