## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 348: 54

#### Answer

$\frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$

#### Work Step by Step

The standard form of the equation of the elipse when the major axis is: - horizontal: $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ in which $(h,k)$ is the center and $2a$ is the major axis length and $2b$ is the minor axis length. - vertical: $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ in which $(h,k)$ is the center and $2a$ is the major axis length and $2b$ is the minor axis length. Vertices: $(-2,3)~~and~~(6,3)$ The center is the midpoint: $\frac{(-2,3)+(6,3)}{2}=(2,3)$ The elipse is in the horizontal position. The distance between the vertices is equal to $2a$: $2a=6-(-2)=6+2=8$ $a=4$ $2b=6$ $b=3$ $\frac{(x-2)^2}{4^2}+\frac{(y-3)^2}{3^2}=1$ $\frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$

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