what is the foci of the ellipse
(x+5)^2/4+(y-1)^2/16=1
it's just like the last problem. however, this ellipse is vertical because b > a. the foci are at (h, k +/- c).
h = -5
k = 1
a = 4
b = 2
4^2 = 2^2 + c^2
16 = 4 + c^2
12 = c^2
2 x the square root of 3 = c
foci = (-5, 1 +/- (2 x the square root of 3))