Find the equation of Hyperbola satisfying the following conditions: Vertices `(0,pm5)`, foci `(0,pm8)`

Explanation

Vertices `-=(0,pm5)` and focis `-=(0,pm8)`
`rArr` The transverse axis of hyperbola is along y-axis, `:.b=5" "and" "be=8`
Now, `a^(2)=b^(2)(e^(2)-1)`
`=b^(2)e^(2)-b^(2)=8^(2)-5^(2)=39`
and equation of hyperbola
`(y^(2))/(5^(2))-(x^(2))/(a^(2))=1`
`rArr(y^(2))/(5^(2))-(x^(2))/(a^(2))=1rArr(y^(2))/(25)-(x^(2))/(39)=1`