SOLUTION: The foci of a hyperbola x^2/16 -(y-2)^2/9=1

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Question 83473: The foci of a hyperbola
x^2/16 -(y-2)^2/9=1
Answer by akhilreddy90(1) About Me  (Show Source):
You can put this solution on YOUR website!
using the formulae for eccentricity we get e=4/5
then for standard hyperbola foci r =(ae,o) or (-ae,o)
but for this one, X=ae
i.e,x=4*4/5=4
thenY=o,but y=2
therfore,foci r (5,2),(-5,2)