Here, foci `(0,pm13)` lie on y-axis
`:." "be=13rArrb^(2)e^(2)=169`
and conjugate axis 2a=24
`:." "a=12rArra^(2)=144`
Now, `a^(2)=b^(2)(e^(2)-1)`
`rArr" "a^(2)=b^(2)e^(2)-b^(2)`
`rArr" "144=169-b^(2)`
`rArr" "b^(2)=25`
`:.` Equation of hyperbola
`(y^(2))/(b^(2))-(x^(2))/(a^(2))=1rArr(y^(2))/(25)-(x^(2))/(144)=1`