Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 348: 42a


$x^2 =-\dfrac{25}{4} (y-48)$

Work Step by Step

We need to write the Standard form of a parabola with a vertical axis as follows: $(x-h)^2 =4p (y-k)$ Now, $(x-0)^2 =4p (y-48)$ $h=0$ and $k=58$ Vertex: $(0, 48)$ Points are: $(10 \sqrt 3,0)$ $(10\sqrt 3-0)^2 =4p (0-48) \implies p=-\dfrac{25}{16}$ Back substitution: $(x-0)^2 =4(-\dfrac{25}{16}) (y-48)$ Our answer is: $x^2 =-\dfrac{25}{4} (y-48)$
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