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 \( \sqrt{2}x - y + 4\sqrt{2}k = 0 \) and \( \sqrt{2}kx + ky - 4\sqrt{2} = 0 \), we will follow these steps:

### Step 1: Solve for \( k \) from the first equation
From the first equation:
\[ \sqrt{2}x - y + 4\sqrt{2}k = 0 \]
Rearranging gives:
\[ 4\sqrt{2}k = y - \sqrt{2}x \]
Thus,
\[ k = \frac{y - \sqrt{2}x}{4\sqrt{2}} \]

### Step 2: Substitute \( k \) into the second equation
Now, we substitute \( k \) into the second equation:
\[ \sqrt{2}kx + ky - 4\sqrt{2} = 0 \]
Substituting \( k \):
\[ \sqrt{2}\left(\frac{y - \sqrt{2}x}{4\sqrt{2}}\right)x + \left(\frac{y - \sqrt{2}x}{4\sqrt{2}}\right)y - 4\sqrt{2} = 0 \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{1}{4}\left(y - \sqrt{2}x\right)x + \frac{1}{4}\left(y - \sqrt{2}x\right)y - 4\sqrt{2} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ (y - \sqrt{2}x)x + (y - \sqrt{2}x)y - 16\sqrt{2} = 0 \]
Expanding this gives:
\[ yx - \sqrt{2}x^2 + y^2 - \sqrt{2}xy - 16\sqrt{2} = 0 \]
Combining like terms:
\[ y^2 - \sqrt{2}xy + yx - \sqrt{2}x^2 - 16\sqrt{2} = 0 \]
This simplifies to:
\[ y^2 - 2\sqrt{2}xy - \sqrt{2}x^2 - 16\sqrt{2} = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ y^2 - 2\sqrt{2}xy - \sqrt{2}x^2 = 16\sqrt{2} \]

### Step 5: Identify the conic section
To identify the conic section, we can rewrite the equation:
\[ \frac{y^2}{32} - \frac{x^2}{16} = 1 \]
This is in the form of a hyperbola:
\[ \frac{y^2}{32} - \frac{x^2}{16} = 1 \]

### Conclusion
The locus of the point of intersection of the given lines is a hyperbola.