1. Let the point of intersection of the two lines be $(x,y)$.

The lines are: $\sqrt{2}x-y+4\sqrt{2}k=0 \quad ...(1)$ $\sqrt{2}kx+ky-4\sqrt{2}=0 \quad ...(2)$ with $k\neq 0$.

We must eliminate the parameter $k$ to get the locus.

2. Express $k$ from each equation.

From (1): $\sqrt{2}x-y=-4\sqrt{2}k$ $k=\frac{y-\sqrt{2}x}{4\sqrt{2}}$

From (2): $k(\sqrt{2}x+y)=4\sqrt{2}$ $k=\frac{4\sqrt{2}}{\sqrt{2}x+y}$

Equating the two expressions for $k$: $\frac{y-\sqrt{2}x}{4\sqrt{2}}=\frac{4\sqrt{2}}{\sqrt{2}x+y}$

3. Simplify the equation.

Cross-multiplying, $(y-\sqrt{2}x)(\sqrt{2}x+y)=32$

Now use the identity: $(a-b)(a+b)=a^2-b^2$ with $a=y$, $b=\sqrt{2}x$.

So, $y^2-(\sqrt{2}x)^2=32$ $y^2-2x^2=32$

Divide by $32$: $\frac{y^2}{32}-\frac{x^2}{16}=1$

4. Identify the conic.

This is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ which is a hyperbola.

Here, $a^2=32,\quad b^2=16$ $a=4\sqrt{2},\quad b=4$

5. Check the given properties.

For the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ its eccentricity is $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{32}}=\sqrt{\frac32}=\frac{\sqrt6}{2}$ which is not $\sqrt3$.

So option C is false.

The length of transverse axis is $2a=2(4\sqrt2)=8\sqrt2$ So option D is true.

6. Final conclusion.

The locus is a hyperbola with transverse axis length $8\sqrt{2}$.

Therefore, the correct option is: $\boxed{\text{D}}$

7. Comparison with stored answer.

Stored correct answer: D

Our derived answer: D

They agree.