Answer option A
From the given graph is a Vertical ellipse
Center of ellipse = (-2,-3)
Vertices are (-2,3) and (-2,-9)
Co vertices are (-6,-3) and (2,-3)
The distance between center and vertices = 6, so a= 6
The distance between center and covertices = 4 , so b= 4
The general equation of vertical ellipse is

(h,k) is the center
we know center is (-2,-3)
h= -2, k = -3 , a= 6 and b = 4
The standard equation becomes


Foci are (h,k+c) and (h,k-c)

Plug in the a=6 and b=4


, we know h=-2 and k=-3
Foci are
and 
Option A is correct