## College Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Conic Sections - Exercise Set 7.1 - Page 671: 28

#### Answer

$\displaystyle \frac{x^{2}}{7}+\frac{y^{2}}{16}=1$

#### Work Step by Step

With the given information, the foci are $c=3$ units above/below the point (0,0) the major axis is vertical, $a=4$ so the standard form the equation is $\displaystyle \frac{(x-h)^{2}}{b^{2}}+ \displaystyle \frac{(y-k)^{2}}{a^{2}}=1$ $(h,k)=(0,0)$ $a=4$ $c=3$ From $b^{2}=a^{2}-c^{2}$ $b^{2}=16-9=7$ so the equation is $\displaystyle \frac{x^{2}}{7}+\frac{y^{2}}{16}=1$

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