Here foci are (±4, 0) which lie on x-axis. So the equation of the hyperbola in standard form is $\frac{\frac{x^2}{a^{-y^2}}}{b^2}$ = 1. Now Foci are (±4, 0) ⇒ c = 4. Length of latus rectum is $\frac{2b^2}{a}$ = 12 ⇒ $b^2$ = 6a We know that $c^2$ = $a^2 + b^2$ ∴ $(4)^2$ = $a^2$ + 6a ⇒ $a^2$ + 6a – 16 = 0 ⇒ (a + 8)(a – 2) = 0 ⇒ a = 2 (Since a = – 8 is not possible) ∴ $a^2$ = 4 ⇒ $b^2$ = 6 × 2 = 12 Thus, required equation of hyperbola is $\frac{x^2}{4} - \frac{y^2}{12}$ = 1.