Vertices `(0,pm3)` foci `(0,pm5)`
`:.` The transverse axis of hyperbola is along y-axis.
`:.b=3andbe=5`
`rArrb^(2)=9andb^(2)e^(2)=25`
`:.a^(2)e^(2)-b^(2)=25-9-16`
and equation of hyperbola
`(y^(2))/(b^(2))-(x^(2))/(a^(2))=1rArr(y^(2))/(9)-(x^(2))/(16)=1`
