Find the equations of the hyperbola satisfying the given conditions :Foci $\left(\pm 4,0\right)$, the latus rectum is of length 12

Solution: Here, Foci of hyperbola $=\left(\pm 4,0\right)$
That means the transverse axis of the hyperbola is $X$-axis.
So, the equation will be of the type,
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\to \left(1\right)$
Also, $c=4$
Length of latus rectum $=12$
$\therefore 2\frac{b^2}{a}=12\Rightarrow b^2=6a$
In a hyperbola, $c^2=a^2+b^2$
Putting value of $c$ and $b^2$,
$\therefore 4^2=a^2+6a$
$\Rightarrow a^2+6a-16=0$
$\Rightarrow a^2+8a-2a-16=0$
$\Rightarrow \left(a+8\right)\left(a-2\right)=0$
$\therefore a=-8\text{and}a=2$
But, $a$ can not be negative, so,$a=2\Rightarrow a^2=4$
$b^2=6a=6\ast 2=12$
Putting values of $a^2$ and $b^2$ in (1),
$\frac{x^2}{4}-\frac{y^2}{12}=1$