## Algebra 2 (1st Edition)

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