Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 708: 5

Answer

Vertex: : $(3,-1)$ Focus: $(7,-1)$ Directrix: $x=-1$

Work Step by Step

Recall: The parabola $(y-b)^2=4p(x-a)$ has a vertex $(a,b)$, a focus $(a+p,b)$, and a directrix $x=-p+a$. We have $(y+1)^2=16(x-3)$ or equivalently $(y-(-1))^2=16(x-3)$. Then, $(a,b)=(3,-1)$ $4p=16$ $p=\frac{16}{4}$ $p=4$ So, the vertex is $(3,-1)$, the focus is $(7,-1)$, and the directrix is $x=-1$. Sketch the graph:
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