The locus of the point of intersection of the lines, `sqrt(2)x-y+4sqrt(2)k=0" and "sqrt(2)kx+ky-4sqrt(2)=0` (k is any non-zero real parameter), is:

Options

A. A hyperbola with length of its transverse axis `8sqrt(2)`
B. A hyperbola whose eccentricity is `sqrt(3)`
C. An ellipse whose eccentricity is `(1)/(sqrt(3))`
D. An ellipse with length of its major axis `8sqrt(2)`

Answer

To find the locus of the point of intersection of the lines given by the equations:

1. \( \sqrt{2}x - y + 4\sqrt{2}k = 0 \)
2. \( \sqrt{2}kx + ky - 4\sqrt{2} = 0 \)

we can follow these steps:

### Step 1: Rearranging the first equation
From the first equation, we can express \( y \) in terms of \( x \) and \( k \):

\[ y = \sqrt{2}x + 4\sqrt{2}k \]

### Step 2: Rearranging the second equation
From the second equation, we can express \( k \) in terms of \( x \) and \( y \):

\[ \sqrt{2}kx + ky = 4\sqrt{2} \]

Factoring out \( k \):

\[ k(\sqrt{2}x + y) = 4\sqrt{2} \]

Thus,

\[ k = \frac{4\sqrt{2}}{\sqrt{2}x + y} \]

### Step 3: Substituting \( k \) into the first equation
Now substitute the expression for \( k \) from Step 2 into the expression for \( y \) in Step 1:

\[ y = \sqrt{2}x + 4\sqrt{2} \left(\frac{4\sqrt{2}}{\sqrt{2}x + y}\right) \]

### Step 4: Simplifying the equation
Multiply through by \( \sqrt{2}x + y \) to eliminate the fraction:

\[ y(\sqrt{2}x + y) = \sqrt{2}x(\sqrt{2}x + y) + 16 \]

Expanding both sides:

\[ y\sqrt{2}x + y^2 = 2x^2 + \sqrt{2}xy + 16 \]

Rearranging gives:

\[ y^2 + (y\sqrt{2}x - \sqrt{2}xy) - 2x^2 - 16 = 0 \]

### Step 5: Collecting like terms
This simplifies to:

\[ y^2 - 2x^2 - 16 = 0 \]

### Step 6: Rearranging into standard form
Rearranging gives us:

\[ \frac{y^2}{16} - \frac{x^2}{8} = 1 \]

This is the equation of a hyperbola in standard form.

### Conclusion
The locus of the point of intersection of the given lines is:

\[ \frac{y^2}{16} - \frac{x^2}{8} = 1 \]