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: 25


$y^{2}=-8x$ The latus rectum has endpoints $(-2,-4)$ and $(-2,4)$

Work Step by Step

1. The directrix is vertical$\quad \Rightarrow\quad $the axis of symmetry is horizontal. 2.$\quad $The focus lies on the line $y=0$. So does the vertex. 3.$\quad $The focus is left of the directrix, the parabola opens left. $a=2$ 4.$\quad $From Table 2, $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h-a,k)}&{x=h+a}&{(y-k)^{2}=-4a(x-h)} &left \\ \hline \end{array}$ $(h-a,k)=(-2,0)\quad \Rightarrow$Vertex: $(0,0)$ The equation is $\quad (y-0)^{2}=-4(2)(x-0)$ that is, $\quad y^{2}=-8x$ 5. For $x=-2, \quad y^{2}=16\quad \Rightarrow\quad y=\pm 4$ The latus rectum (line segment parallel to the directrix, containing the focus) has endpoints $(-2,-4)$ and $(-2,4)$ We have enough details for the graph.
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