Precalculus (6th Edition)

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

Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1002: 56

Answer

$$\frac{{{x^2}}}{{72}} - \frac{{{y^2}}}{{72}} = 1$$

Work Step by Step

$$\eqalign{ & {\text{hyperbola with foci at }}\left( {0, \pm 12} \right){\text{ and asymptotes }}y = \pm x \cr & {\text{The coordinates of the foci are }}\left( {0, \pm c} \right),{\text{ then}} \cr & {\text{The equation of the ellipse is of the form }}\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1 \cr & {\text{foci}}\left( {0, \pm c} \right),\,\,\left( {0, \pm 12} \right) \to c = 12 \cr & {\text{Asymptotes }}y = \pm \frac{b}{a}x \cr & \frac{b}{a} = 1,\,\,\,a = b \cr & {c^2} = {a^2} + {b^2} \cr & {c^2} = 2{a^2} \cr & {12^2} = 2{a^2} \cr & {a^2} = 72 = {b^2} \cr & \cr & {\text{The equation of the hyperbola is}} \cr & \frac{{{x^2}}}{{72}} - \frac{{{y^2}}}{{72}} = 1 \cr} $$
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