#### 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.