Precalculus (6th Edition)

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

Chapter 10 - Analytic Geometry - 10.3 Hyperbolas - 10.3 Exercises - Page 988: 24

Answer

$$\eqalign{ & {\text{Center }}\left( {1, - 5} \right) \cr & {\text{Vertices}}:\left( {1, - 7} \right){\text{ and }}\left( {1, - 3} \right) \cr & {\text{Foci: }}\left( {1, - 15} \right){\text{ and }}\left( {1,5} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{1}{2}\left( {x - 1} \right) - 5 \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{range: }}\left( { - \infty , - 7} \right] \cup \left[ { - 3,\infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & \frac{{{{\left( {y + 5} \right)}^2}}}{4} - \frac{{{{\left( {x - 1} \right)}^2}}}{{16}} = 1 \cr & {\text{The equation is written in the form }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {y + 5} \right)}^2}}}{4} - \frac{{{{\left( {x - 1} \right)}^2}}}{{16}} = 1,{\text{ then }}a = 2,\,\,b = 4,\,\,\,h = 1,\,\,\,k = - 5 \cr & c = \sqrt {36 + 64} = 10 \cr & \cr & {\text{Characteristics:}} \cr & {\text{Center: }}\left( {h,k} \right) \cr & {\text{Center: }}\left( {1, - 5} \right) \cr & {\text{Transverse axis: horizontal}} \cr & {\text{vertices: }}\left( {h,k \pm a} \right) \cr & {\text{vertices: }}\left( {1, - 5 \pm 2} \right) \cr & {\text{vertices: }}\left( {1, - 7} \right){\text{ and }}\left( {1, - 3} \right) \cr & {\text{foci:}}\left( {h,k \pm c} \right) \cr & {\text{foci:}}\left( {1, - 5 \pm 10} \right) \cr & {\text{foci:}}\left( {1, - 15} \right){\text{ and }}\left( {1,5} \right) \cr & {\text{asymptotes: }}y = \pm \frac{a}{b}\left( {x - h} \right) + k \cr & {\text{asymptotes: }}y = \pm \frac{1}{2}\left( {x - 1} \right) - 5 \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{The range is: }}\left( { - \infty ,k - a} \right] \cup \left[ {k + a,\infty } \right) \cr & {\text{The range is: }}\left( { - \infty , - 7} \right] \cup \left[ { - 3,\infty } \right) \cr} $$
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