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: 51

Answer

$${\left( {y - 2} \right)^2} = 12x$$

Work Step by Step

$$\eqalign{ & {\text{parabola with focus at }}\left( {3,2} \right){\text{ and directrix }}x = - 3 \cr & {\text{The directrix is }}x = - p + h,{\text{ then }} \cr & {\text{the equation of the parabola is of the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr & \cr & {\text{Focus }}\left( {p + h,k} \right):\left( {3,2} \right),\,\,\,\,\,\,p + h = 3,\,\,\,\,\,k = 2 \cr & x = - p + h,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - p + h = - 3 \cr & {\text{Solving these equations we obtain:}} \cr & p = 3{\text{ and }}h = 0 \cr & \cr & {\text{The equation of the parabola is}} \cr & {\left( {y - 2} \right)^2} = 4\left( 3 \right)\left( {x - 0} \right) \cr & {\left( {y - 2} \right)^2} = 12x \cr} $$
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