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 635: 6


The vertices are $(3,-1\pm7)$ and $(3\pm 4,-1)$ . The foci are $(3,-1\pm \sqrt{33})$.

Work Step by Step

The equation $$ \left(\frac{x-3}{4}\right)^{2}+\left(\frac{y+1}{7}\right)^{2}=1 $$ is an ellipse with center $(3,-1)$. We have $a=4$, $b=7$, $c=\sqrt{b^2-a^2}=\sqrt{33}$ and hence: When the center is at the origin, the foci are $(0,\pm \sqrt{33})$ and the vertices are $(\pm 4,0)$ and $(0.\pm 7)$ Then the translation by $(3,-1)$ gives: The new vertices are $(3,-1\pm7)$ and $(3\pm 4,-1)$ . The new foci are $(3,-1\pm \sqrt{33})$.
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