## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Mixed Review - Page 858: 51

#### Answer

$\dfrac{(y-2)^2}{36}-\dfrac{(x+3)^2}{144}=1$; Hyperbola.

#### Work Step by Step

Given: $x^2-4y^2+6x+16y+137=0$ $(x+3)^2-4(y-2)^2+144=0$ This gives: $4(y-2)^2-(x+3)^2=144$ Thus, we have $\dfrac{(y-2)^2}{36}-\dfrac{(x+3)^2}{144}=1$ This is the standard form of a Hyperbola.

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