Explanation

(c)
Given equation of hyperbola can be rewritten as
$\frac{x^{2}}{\left( \frac{12}{5} \right)^{2}} - \frac{y^{2}}{\left( \frac{9}{5} \right)^{2}} = 1$
$\therefore Eccentricity\ given\ by\ e^{'2} = 1 + \frac{b^{'2}}{a^{'2}}$
$\Rightarrow e^{'2} = 1 + \frac{9}{16} = \frac{25}{16} \Rightarrow e^{'} = \frac{5}{4}$
The foci of a hyperbola are
$\left( \pm a^{'}e^{'},\ 0 \right) = \left( \pm \frac{12}{5} \times \frac{5}{4},\ 0 \right) = ( \pm 3,\ 0)$
Given equation of ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1$
Foci of an ellipse are $( \pm ae,\ o) = ( \pm 4e,\ o)$. But given focus of ellipse and
hyperbola coincide, then
$4e = 3\ \Rightarrow \ e = \frac{3}{4}$
Also, $b^{2} = a^{2}\left( 1 - e^{2} \right)$
$= 16\left( 1 - \frac{9}{16} \right) = 16 - 9 = 7$