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: 36

Answer

$(y-4)^{2}=12(x+1)$ The latus rectum has endpoints $(2,-2)$ and $(2,10).$

Work Step by Step

1.$\quad $The directrix is vertical, so the parabola opens left or right. The focus and the vertex lie on a horizontal line, $y=4$. The focus is 6 units right of the directrix, so it opens right. The vertex is halfway between the focus and directrix, 3 units from either, $(h,k)=(-1,4),\ \quad a=3.$ 2.$\quad $ Using table 2: $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$ The equation is $(y-4)^{2}=4(3)(x+1)$ that is, $\quad (y-4)^{2}=12(x+1)$ 3.$\quad $For the latus rectum (line segment parallel to the directrix, containing the focus), set $x=2$ $(y-4)^{2}=12(2+1)$ $(y-4)^{2}=36$ $y-4=\pm 6$ $y=4\pm 6$ The latus rectum has endpoints $(2,-2)$ and $(2,10).$
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