D?

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If the distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$, then the equation of the hyperbola is

- $y^2-x^2=32$
- $x^2-y^2=16$
- $y^2-x^2=16$
- $x^2-y^2=32$

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__ 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.