find the equation of hyperabola where foci are (0,12) and (0,-12)and the length of the latus rectum is 36

Explanation

The focus of the hyperbola lies on y-axis, terefore, let the equation of the hyperbola is
`(y^(2))/(b^(2))-(x^(2))/(a^(2))=1`
Co-ordinates of foci `=(0,pmbe)`
be=12
and latus rectum `(2a^(2))/(b)=36`
`rArr" "a^(2)=18b`
`rArr" "b^(2)(e^(2)-1)=18b`
`rArr" "144-b^(2)=18b`
`rArr""b^(2)+18b-144=0`
`rArr" "(b+24)(b-6)=0`
`rArr" "b=-24orb=6`
b=-24
(which is not possible) `:.` b=6
Therefore, equation of hyerbola
`(y^(2))/(36)-(x^(2))/(108)=1`
`rArr" "3y^(2)-x^(2)=108`.