Question

Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,0)$ the latus rectum is of length $12$

Answer 1

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 $(\pm 4,0)\Rightarrow ae=c=4$
Length of latus rectum $=12$
$\Rightarrow \frac{2b^2}{a}=12$
$\Rightarrow$ $b^2$ $=6a$
We know that $a^2+b^2=c^2$
$\therefore a^2+6a=16$
$\Rightarrow a^2+6a-16=0$
$\Rightarrow a^2+8a-2a-16=0$
$\Rightarrow (a+8)(a-2)=0$
$\Rightarrow a=-8,2$
Since a is non-negative $a=2$
$\therefore b^2=6a=6\times 2=12$
Thus the equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$