Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0),

Solution:

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

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

$\Rightarrow \frac{2 b^{2}}{a}=12$

$\Rightarrow b^{2}=6 a$

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore a^{2}+6 a=16$

$\Rightarrow a^{2}+6 a-16=0$

$\Rightarrow a^{2}+8 a-2 a-16=0$

$\Rightarrow(a+8)(a-2)=0$

$\Rightarrow a=-8,2$

Since a is non-negative, a = 2.

$\therefore b^{2}=6 a=6 \times 2=12$

Thus, the equation of the hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.