Explanation

(c)
Given length of latusrectum $= \ \frac{2b^{2}}{a} = 9\$
$\Rightarrow b^{2} = \frac{9a}{2}$ …(i)
and $e = \frac{5}{4}$
$\frac{25}{16} = 1 + \frac{b^{2}}{a^{2}}$
$\Rightarrow \ \ 1 + \frac{9a}{2a^{2}} = \frac{25}{16}\$ [form eq.(i)]
$\Rightarrow \frac{9}{2a} = \frac{9}{16}\ \Rightarrow \ a = 8$
On putting the value of $a$ in Eq. (i), we get
$b^{2} = \frac{9 \times 8}{2}\ \Rightarrow \ b = 6$
$\therefore$ Equation of hyperbola is
$\frac{x^{2}}{8^{2}} - \frac{y^{2}}{6^{2}} = 1\ \Rightarrow \ \frac{x^{2}}{64} - \frac{y^{2}}{36} = 1$