Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.3 - The Parabola - Exercise Set - Page 997: 35



Work Step by Step

General form of a horizontal parabola is given as: $(x-h)^2=4p(y-k)$...(1) Here, $\text{Vertex}=(h,k)$ and focus is: $(h, k+p)$ As we are given $(x-2)^2=8(y-1)$ From equation (1), we get: $h=2,k=1$ and $4p=8\implies p=2$ $\text{Vertex}=(h,k) \implies (1,1)$ Also, focus is: $(h,k+p) \implies f(2,1+2)=f(2,3)$ Directrix is: $y=k-p \implies y=1-2=-1$ Hence, Directrix is: $y=-1$
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.