Answer: $\mathbf D$
Explanation:
General equation of hyperbola $ = \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Ecentricity is given by $e = \frac{\sqrt{a^2-b^2}}{a}$
Focal distance$=16$
$\Rightarrow 2ae = 16$
$\Rightarrow ae = 8$
Now, $e = \sqrt 2$
$\therefore a\sqrt2 = 8$
$a = \frac{8}{\sqrt 2} = \frac{8}{\sqrt 2} \times \frac{\sqrt 2}{\sqrt 2} = 4 \sqrt 2$
$\therefore a^2 = 32$
$e^2 = \frac{a^2 - b^2}{a^2}$
$\Rightarrow b^2 = a^2e^2 -a^2$
$\Rightarrow b^2 = 64 -32 = 32$
So, the equation of the hyperbola $= \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
$\therefore \mathbf D$ is the correct option.