# Chapter 9 - Section 9.1 - The Ellipse - Exercise Set - Page 966: 36

$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{1}=1$

#### Work Step by Step

Step 1. Based on the given conditions, we have $a=\frac{8-2}{2}=3, b=\frac{3-1}{2}=1$ with a horizontal major axis. The ellipse is centered at $(\frac{2+8}{2},\frac{1+3}{2})$ or $(5,2)$. Step 2. We can write the equation as $\frac{(x-5)^2}{9}+\frac{(y-2)^2}{1}=1$

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