Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 9 - Analytic Geometry - Section 9.2 The Parabola - 9.2 Assess Your Understanding - Page 667: 29

Answer

$(x-2)^2=-8(y+3)$, $(-2,-5),(6,-5)$. See graph

Work Step by Step

1. Given vertex $(2,-3)$ and focus $(2,-5)$ (opens down), we have $p=-3+5=2$ and equation (as in a standard form) $(x-2)^2=-4(2)(y+3)$ or $(x-2)^2=-8(y+3)$ 2. Let $y=-5$, we have $(x-2)^2=-8(-5+3)=16$ and $x=2\pm4$, thus we have the two points that define the latus rectum $(-2,-5),(6,-5)$ 3. See graph.
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