## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Review Exercises - Page 1035: 10

#### Answer

$\frac{x^2}{27}+\frac{y^2}{36}=1$

#### Work Step by Step

Step 1. With the given conditions, we have $c=3, a=6$; thus $b^2=6^2-3^2=27$, with foci on the y-axis. Step 2. The center is at $(0,0)$, so we can write the equation as $\frac{x^2}{27}+\frac{y^2}{36}=1$

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