Given:

Foci (± 4, 0) and the latus rectum is of length 12

Here, the foci are on x-axis.

The equation of the hyperbola is of the form \(\dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1\)

Since, the foci are (± 4, 0), so, c = 4

Length of latus rectum is 12

\(\dfrac{ 2b^2}{a}\) = 12

2b^{2} = 12a

b^{2} = \( \dfrac{12a}{2}\)

= 6a

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 6a = 16

a^{2} + 6a – 16 = 0

a^{2} + 8a – 2a – 16 = 0

(a + 8) (a – 2) = 0

a = -8 or 2

Since, a is non – negative, a = 2

So, b^{2} = 6a

= 6 × 2

= 12

The equation of the hyperbola is \(\dfrac{ x^2}{4} – \dfrac{y^2}{12} = 1\)

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