Explanation

(d)
Given, $S\ (6,\ 4)$ and $S^{'}( - 4,\ 4)$ and eccentricity, $e = 2$
$\therefore\ \ SS^{'} = \sqrt{(6 + 4)^{2} + (4 - 4)^{2}} = 10$
But $SS^{'} = 2\ ae$
$\therefore\ \ 2a \times 2 = 10$
$\Rightarrow a = \frac{5}{2}$
And we know that,
$b^{2} = a^{2}(e^{2} - 1)$
$\Rightarrow \ \ b^{2} = \frac{25}{4}(4 - 1) = \frac{75}{4}$
Centre of hyperbola is $\left( \frac{6 + ( - 4)}{2},\frac{4 + 4}{2} \right) = (1,\ 4)$
$\therefore\$ Equation of hyperbola is $\frac{(x - 1)^{2}}{\frac{25}{4}} - \frac{(y - 4)^{2}}{\frac{75}{4}} = 1$
$\Rightarrow \ \ \frac{4(x - 1)^{2}\ }{25} - \frac{4(y - 4)^{2}}{75} = 1$