Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.3 - The Parabola - Exercise Set - Page 999: 98


focus $(0,-\frac{E}{4A})$, directrix $y=\frac{E}{4A}$

Work Step by Step

Step 1. Rewriting the equation as $x^2=-\frac{E}{A}y$, we have $4p=-\frac{E}{A}$ and $p=-\frac{E}{4A}$ Step 2. The vertex is at $(0,0)$. We can identify the focus at $(0,-\frac{E}{4A})$ with directrix $y=\frac{E}{4A}$
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