Question

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 3\sqrt{5},0)$ the latus rectum is of length $8$

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 $(\pm 3\sqrt{5},0)\Rightarrow ae=c=\pm 3\sqrt{5}$
Length of latus rectum $=8$
$\Rightarrow \frac{2b^2}{a}=8$
$\Rightarrow$ $b^2=4a$
We know that $a^2+b^2=c^2$
$\therefore a^2+4a=45$
$\Rightarrow a^2+4a-45=0$
$\Rightarrow a^2+9a-5a-45=0$
$\Rightarrow (a+9)(a-5)=0$
$\Rightarrow a=-9,5$
Since a in non-negative $a=5$
$\therefore b^2=4a=4\times 5=20$
Thus the equation of the hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}=1$