Precalculus (6th Edition)

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

Chapter 10 - Analytic Geometry - 10.1 Parabolas - 10.1 Exercises - Page 967: 43

Answer

$${\left( {x - 4} \right)^2} = 8\left( {y - 3} \right)$$

Work Step by Step

$$\eqalign{ & {\text{vertex }}\left( {4,3} \right),\,\,{\text{focus }}\left( {4,5} \right) \cr & {\text{Because the focus is upward the vertex}} \cr & {\text{The equation is of the form }}{\left( {x - h} \right)^2} = 4p\left( {y - k} \right) \cr & {\text{With vertex }}\left( {h,k} \right){\text{ and focus }}\left( {h,p + k} \right) \cr & h = 4,\,\,k = 3,\,\,\,\,p + 3 = 5,\,\,\,p = 2 \cr & \cr & {\text{The equation is }} \cr & {\left( {x - h} \right)^2} = 4p\left( {y - k} \right) \cr & {\left( {x - 4} \right)^2} = 4\left( 2 \right)\left( {y - 3} \right) \cr & {\left( {x - 4} \right)^2} = 8\left( {y - 3} \right) \cr} $$
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