Find the equation of the hyperbola where foci are (0,±12) and the length of the latus rectum is 36.

Since foci are (0,±12), it follows that c = 12.

Length of the latus rectum =2b²/a = 36 ⇒ b² = 18a

∴ c² = a² + b²

⇒ 144 = a² + 18a

i.e., a² + 18a - 144 = 0

So a = -24, 6

Since a cannot be negative, we take a = 6 and so, b² = 108.

∴ The equation of the required hyperbola is y²/36 - x²/108 = 1