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: 34


$\text{Vertex}= (1,1)$ Focus: $f(0,1)$ Directrix is: $x=2$ 'Graph b'

Work Step by Step

General form of a horizontal parabola is given as: $(y-k)^2=4p(x-h)$...(1) Here, $\text{Vertex}=(h,k)$ and focus is: $(h+p, k)$ As we are given $(y-1)^2=-4(x-1)$ From equation (1), we get: $k=1, h=1$ and $4p=-4 \implies p=-1$ $\text{Vertex}=(h,k) \implies (1,1)$ Also, focus is: $(h+p,k) \implies f(1-1,1)=f(0,1)$ Directrix is: $x=h-p \implies x=1+1=2$ Hence, $\text{Vertex}= (1,1)$ Focus: $f(0,1)$ Directrix is: $x=2$ Thus, the 'Graph b' matches with all the above conditions.
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