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# Find the equations of the hyperbola satisfying the given conditions: Foci: $(\pm 4 0)$, Latus Rectum is of length $12$.

Toolbox:
Given Foci: $(\pm 4, 0) \rightarrow c = 4$.
Given Latus Rectum is of length $12 \rightarrow 112 = \large\frac{2b^2}{a}$$\rightarrow b^2 = 6a Since a^2+b^2 = c^2 \rightarrow a^2+6a=4\rightarrow a = 2 (We discard a=-8 since a is non-negative). If a = 2 \rightarrow b^2 = 6 \times 2 = 12 Therefore, the equation of the hyperbola is as follows: \;\large\frac{x^2}{4}$$-\large\frac{y^2}{12}$$=1$