## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 349: 94b

#### Answer

$\dfrac{(x-5)^2}{16}+\dfrac{(y-6)^2}{25}=1$

#### Work Step by Step

The standard form of the equation of the ellipse when the major axis is horizontal can be expressed as: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Where $(h,k)$ is the center, $2a$ is the major axis length, and $2b$ is the minor axis length. The standard form of the equation of the ellipse when the major axis is vertical can be expressed as: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Where $(h,k)$ is the center, $2a$ is the major axis length, and $2b$ is the minor axis length. The center $(h,k): (5,6)$ The vertices are located at the end points of the vertical major axis $(5,1), (5,11)$. The major axis length is $2a =10 \implies a=5$ and the minor axis length is $2b=8 \implies b=4$ Now, $c=\sqrt {a^2-b^2}=\sqrt {5^2-4^2}=3$ Foci: $(5,3), (5,9)$ The major axis is vertical, so we have $a=5, b=4$: $\dfrac{(x-5)^2}{16}+\dfrac{(y-6)^2}{25}=1$

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