Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 35

Answer

$(y+1)^2=\frac{1}{2}(x-3)$

Work Step by Step

Parabola passing through $(-15,2)$ Put $x=-15$ and $y=2$ Therefore, $(2+1)^2=4p(-15-3)$ $p=-\frac{1}{8}$ Thus, $(y+1)^2=\frac{1}{2}(x-3)$
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