Question 454238
{{{(x+6)^2/12 + (y-4)^2/16 = 1}}}
Use braces { and }, 3 on each end, to format.
Long axis is "vertical", parallel to the y-axis (16>12)
Center at (-6,4)
Distance from center to foci = sqrt(16 - 12) = 2
Foci at (-6,2) and (-6,6)
-----------------------
Vertices at center +/- sqrt(16) vertically --> (-6,0) & (-6,8)
Vertices at center +/- sqrt(12) horizontally 
--> (-6-sqrt(12),4) & (-6+sqrt(12),4)