what is the foci of the ellipse

(x+5)^2/4+(y-1)^2/16=1

1 answer

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))
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