Here foci are (±4, 0) which lie on x-axis.
So the equation of the hyperbola in standard form is

x2

a

−

y2

b2

= 1.
Now Foci are (±4, 0) ⇒ c = 4.
Length of latus rectum is

2b2

a

= 12 ⇒ b2 = 6a
We know that c2 = a2+b2
∴ (4)2 = a2 + 6a ⇒ a2 + 6a – 16 = 0
⇒ (a + 8)(a – 2) = 0
⇒ a = 2 (Since a = – 8 is not possible)
∴ a2 = 4 ⇒ b2 = 6 × 2 = 12
Thus, required equation of hyperbola is

x2

4

−

y2

12

= 1.