College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Conic Sections - Exercise Set 7.1 - Page 671: 30


$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{8}=1$

Work Step by Step

With the given information, the foci are $c=2$ units above/below the point (0,0) the major axis is vertical, so the standard form the equation is $\displaystyle \frac{(x-h)^{2}}{b^{2}}+ \displaystyle \frac{(y-k)^{2}}{a^{2}}=1$ The x-intercepts give: $b=2$. $(h,k)=(0,0)$ $b=2$ $c=2$ From $c^{2}=a^{2}-b^{2}$ $a^{2}=b^{2}+c^{2}=4+4=8$ so the equation is $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{8}=1$
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