Explanation

(a)
The given equation is $2x^{2} - 3y^{2} = 5,$ it can be rewritten as $\frac{x^{2}}{\frac{5}{2}} - \frac{y^{2}}{\frac{5}{3}} = 1$
Now, $b^{2} = a^{2}(e^{2} - 1)$
$\Rightarrow \frac{5}{3} = \frac{5}{2}(e^{2} - 1)$
$e^{2} = \frac{2}{5}\left( \frac{5}{3} + \frac{5}{2} \right) = \frac{2}{5}\left( \frac{25}{6} \right)$
$\Rightarrow \ \ e = \sqrt{\frac{5}{3}\ }$
$\therefore\$ Foci of hyperbola $= ( \pm ae,0)$
$= \left( \pm \sqrt{\frac{5}{2}\ } \bullet \sqrt{\frac{5}{3}},\ 0 \right) = \left( \pm \frac{5}{\sqrt{6}},\ 0 \right)$