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: 34

Answer

$(x-3)^{2}=-8y$ The latus rectum has endpoints $(-1,-2)$ and $(7,-2).$

Work Step by Step

The focus lies below the vertex, on the line $x=3$, so the parabola opens down. By table 2, $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h,k-a)}&{y=k+a}&{(x-h)^{2}=-4a(y-k)}& down \\ \hline \end{array}$ $(h,k)=(3,0)$ Given the focus is at $(3,-2)=(h,k-a)$, we find $a:$ $-2=k-a$ $a=2.$ The equation is $\quad (x-3)^{2}=-4(2)(y-0),$ that is, $\quad \fbox{$(x-3)^{2}=-8y$},$ For the latus rectum (line segment parallel to the directrix, containing the focus), set $y=-2$ $(x-3)^{2}=-8(-2)$ $(x-3)^{2}=16$ $x-3=\pm 4$ $x=3\pm 4$ The latus rectum has endpoints $(-1,-2)$ and $(7,-2).$ The directrix has equation $\quad y=k+a\quad \Rightarrow\quad y=2$
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