Question

what are the coordinates of the foci of the conic section shown below ((x+2)^2/16 )-((y-3)^2/9)=1

Answer 1

this is a hyperbola..
center at (-2, 3)

c^2 = a^2 + b^2
where a=4, b=3
so c=5.
your foci is c=5 units above and below the center of the hyperbola.

Answer 2

1) Given equation is that of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.

2) For given equation:
center: (-2,3)
Vertices:
a^2=9
a=3
length of transverse axis=2a=6
vertices: (-2,3±a)=(-2,3±3)=(-2,6) and (-2,0)

3) Foci:
b^2=16
b=4
c^2=a^2+b^2=9+16=25
c=5

Answer 3

See the graph. The axis containing the vertices and the Foci is parallel to the x-axis.

Answer 4

(-6,3) and (2,3)

Answer 5

(-6,3) and (2,3) are the vertices.

Answer 6

hmm..
my bad... thanks @eliassaab :)
your foci should be 5 units left/right of the center (-2, 3)