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: 32

Answer

$(y+2)^{2}=8(x-4)$ The latus rectum has endpoints $(6,-6)$ and $(6,2).$

Work Step by Step

1.$\quad $The vertex and focus lie on the line $y=-2\quad \Rightarrow\quad $parabola opens left or right. The focus is right of the vertex$\quad \Rightarrow\quad $opens right. 2.$\quad $From Table 2, $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$ $(h,k)=(4,-2)$ The equation is $\quad (y+2)^{2}=4a(x-4)\quad $ 3.$\quad $Since the focus is at $\quad (h+a, k)$ and we are given the focus is at$\quad (6,-2)$, we find $a$ $h+a=6$ $4+a=6$ $a=2$ So, the equation is$\quad (y+2)^{2}=8(x-4)$ 4. $\quad $For the latus rectum, set $x=6, \quad $ $(y+2)^{2}=8(6-4)$ $(y+2)^{2}=16$ $(y+2)^{2}=16$ $y+2=\pm 4$ $y=2\pm 4$ The latus rectum (line segment parallel to the directrix, containing the focus) has endpoints $(6,-6)$ and $(6,2).$ 5. $\quad $We have enough details for the graph. Graph the directrix, $\quad x=2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.