Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - 9.4 Graph and Write Equations of Ellipses - 9.4 Exercises - Skill Practice - Page 637: 7


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Work Step by Step

Given: $\frac{x^2}{400}+\frac{y^2}{81}=1$ The equation is in standard form. We can see $a=20, b=9$ The denominator of the $x^2-term$ is greater than that of the $y^2-term$, so the major axis is horizontal. The vertices of the ellipse are at $(\pm a,0)=(\pm 20,0)$. The co-vertices are at $(0,\pm b) = (0,\pm 9)$. Find the foci. $c^2=a^2-b^2=20^2-9^2=319$ so $c=\sqrt 319$ The foci are at $(\pm \sqrt 319,0)$.
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