Find the equation of the hyperbola satisfying the give conditions:

Solution:

Foci $(\pm 3 \sqrt{5}, 0)$, the latus rectum is of length 8 .

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 $(\pm 3 \sqrt{5}, 0), c=\pm 3 \sqrt{5}$.

Length of latus rectum $=8$

$\Rightarrow \frac{2 b^{2}}{a}=8$

$\Rightarrow b^{2}=4 a$

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

$\therefore a^{2}+4 a=45$

$\Rightarrow a^{2}+4 a-45=0$

$\Rightarrow a^{2}+9 a-5 a-45=0$

$\Rightarrow(a+9)(a-5)=0$

$\Rightarrow a=-9,5$

Since $a$ is non-negative, $a=5$.

$\therefore b^{2}=4 a=4 \times 5=20$

Thus, the equation of the hyperbola is $\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$.