## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 349: 76

#### Answer

$\dfrac{(x-3)^2}{4}-\dfrac{(y-2)^2}{5}=1$

#### Work Step by Step

The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ and vertices and foci have the form $(\pm a, 0)$ and $(\pm c,0)$. The standard form of the equation of the hyperbola with a vertical transverse axis can be expressed as: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$ and vertices and foci have the form $(0, \pm, a)$ and $(0, \pm c)$. The center is the midpoint of the vertices : $(3,2)$ We have: $c=3; a=2$ $b^2=c^2-a^2=3^2-2^2= 5$ $\dfrac{(x-3)^2}{2^2}-\dfrac{(y-2)^2}{5}=1$ or, $\dfrac{(x-3)^2}{4}-\dfrac{(y-2)^2}{5}=1$

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