Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 10


$$ \left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1. $$

Work Step by Step

The ellipse $$ \left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1 $$ has the center $(8,-4)$. Shifting the center 4 units to down leads to the ellipse $$ \left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1. $$
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