Question 386857
Standard forms of a hyperbola
(x-h)^2/a^2-(y-k)^2/b^2 = 1 (transverse axis horizontal)
(y-k)^2/a^2-(x-h)^2/b^2 = 1 (transverse axis vertical)
coordinates of given vertices show that transverse axis vertical with center of hyperbola at the origin (0,0)
this also means the y-term comes first, and h=0 and k=0

transverse axis, 8=2a
a=4
a^2=16
from given conjugate axis, 14=2b
b=7
b^2=49

ans: equation of hyperbola, y^2/16-x^2/49 =1

see the following graph

{{{ graph( 300, 200, -20, 20, -16, 16, (16x^2/49+16)^.5,-(16x^2/49+16)^.5) }}}