Question 580267
What are the foci of the equation 
(x+4)^2/9 + (y-2)^2/1 = 1
This is an equation of an ellipse with horizontal axis of the standard form:
(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
..
For given equation:
center: (-4,2)
a^2=9
b^2=1
c^2=a^2-b^2=9-1=8
c=√8
Foci=(-4±c,2)=(-4±√8,2)=(-4±2.83,2)=(-6.83,2) and (-1.17,2)