College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 7 - Section 7.2 - The Parabola - 7.2 Assess Your Understanding - Page 515: 39

Answer

vertex: $\quad (0,0) $ focus: $\quad (0,1)$ directrix: $\quad y=-1$

Work Step by Step

Comparing the form of the equation with table 2, $x^{2}=4y \quad $has the form $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h,k+a)}&{y=k-a}&{(x-h)^{2}=4a(y-k)}&up\\ \hline \end{array}$ $(x-0)^{2}=4a(y-0)\quad \Rightarrow\quad h=0, k=0, a=1$ vertex: $\quad (0,0) $ focus: $\quad (h,k+a)=(0,+1)=(0,1)$ directrix: $\quad y=0-1\quad \Rightarrow\quad y=-1$ To graph, find the endpoints of the latus rectum (line segment parallel to the directrix, containing the focus) by setting $y=1$ $x^{2}=4\cdot 1$ $x=\pm 2$ Endpoints of the latus rectum: $(-2,1), (2,1)$
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