College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 7, Conic Sections - Chapter 7 Review - Exercises - Page 584: 38


$$\frac{x^2}{144} +\frac{y^2}{25} = 1$$

Work Step by Step

The center is $(0,0)$ thus we have the form $$x^2/a^2 +y^2/b^2 = 1$$ Where a is the x intercepts and b is the y intercepts. Plugging those in, $$x^2/144 +y^2/25 = 1$$
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