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
2b2 = 12a
b2 = \( \dfrac{12a}{2}\)
= 6a
It is known that, a2 + b2 = c2
a2 + 6a = 16
a2 + 6a – 16 = 0
a2 + 8a – 2a – 16 = 0
(a + 8) (a – 2) = 0
a = -8 or 2
Since, a is non – negative, a = 2
So, b2 = 6a
= 6 × 2
= 12
The equation of the hyperbola is \(\dfrac{ x^2}{4} – \dfrac{y^2}{12} = 1\)
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