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: 25

Answer

$$\eqalign{ & {\text{Center }}\left( { - 5,3} \right) \cr & {\text{Vertices}}:\left( { - \frac{{21}}{4},3} \right){\text{ and }}\left( { - \frac{{19}}{4},3} \right) \cr & {\text{Foci: }}\left( { - 5 \pm \frac{{\sqrt {17} }}{4},3} \right) \cr & {\text{Asymptotes: }}y = \pm 4\left( {x + 5} \right) + 3 \cr & {\text{domain: }}\left( { - \infty , - \frac{{21}}{4}} \right] \cup \left[ { - \frac{{19}}{4},\infty } \right) \cr & {\text{range: }}\left( { - \infty , + \infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & 16{\left( {x + 5} \right)^2} - {\left( {y - 3} \right)^2} = 1 \cr & {\text{Rewrite the equation as}} \cr & \frac{{{{\left( {x + 5} \right)}^2}}}{{{{\left( {1/4} \right)}^2}}} - \frac{{{{\left( {y - 3} \right)}^2}}}{{{{\left( 1 \right)}^2}}} = 1 \cr & {\text{The equation is written in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {x + 5} \right)}^2}}}{{{{\left( {1/4} \right)}^2}}} - \frac{{{{\left( {y - 3} \right)}^2}}}{{{{\left( 1 \right)}^2}}} = 1,{\text{ then }}a = \frac{1}{4},\,\,b = 1,\,\,\,h = - 5,\,\,\,k = 3 \cr & c = \sqrt {1/16 + 1} = \frac{{\sqrt {17} }}{4} \cr & \cr & {\text{Characteristics:}} \cr & {\text{Center: }}\left( {h,k} \right) \cr & {\text{Center: }}\left( { - 5,3} \right) \cr & {\text{Transverse axis: horizontal}} \cr & {\text{vertices: }}\left( {h \pm a,k} \right) \cr & {\text{vertices: }}\left( { - 5 \pm \frac{1}{4},2} \right) \cr & {\text{vertices: }}\left( { - \frac{{21}}{4},3} \right){\text{ and }}\left( { - \frac{{19}}{4},3} \right) \cr & {\text{foci:}}\left( {h \pm c,k} \right) \cr & {\text{foci:}}\left( { - 5 \pm \frac{{\sqrt {17} }}{4},3} \right) \cr & {\text{asymptotes: }}y = \pm \frac{b}{a}\left( {x - h} \right) + k \cr & {\text{asymptotes: }}y = \pm 4\left( {x + 5} \right) + 3 \cr & {\text{The domain is: }}\left( { - \infty ,h - a} \right] \cup \left[ {h + a,\infty } \right) \cr & {\text{domain: }}\left( { - \infty , - \frac{{21}}{4}} \right] \cup \left[ { - \frac{{19}}{4},\infty } \right) \cr & {\text{The range is: }}\left( { - \infty , + \infty } \right) \cr} $$
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