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Step-by-Step Solution
Step 1: Express y in terms of x and k from the first line
The first line given is
$\,\sqrt{2}x \;-\; y \;+\; 4\sqrt{2}\,k \;=\; 0\,$.
Rearrange to find $y$:
$$
y \;=\;\sqrt{2}x + 4\sqrt{2}\,k.
$$
Step 2: Substitute y into the second line
The second line given is
$\,\sqrt{2} \, k \, x \;+\; k \, y \;-\; 4\sqrt{2} \;=\; 0\,$.
Substitute $\,y = \sqrt{2}x + 4\sqrt{2}\,k\,$ into this equation:
$$
\sqrt{2} \, k \, x \;+\; k\bigl(\sqrt{2}x + 4\sqrt{2}\,k\bigr) \;-\; 4\sqrt{2} \;=\; 0.
$$
Step 3: Simplify and solve for x
Expand and simplify:
$$
\sqrt{2} \, k \, x + k\sqrt{2}x + 4 k \sqrt{2}\,k - 4\sqrt{2} \;=\; 0
\quad\Longrightarrow\quad
2\sqrt{2}\,k\,x + 4\sqrt{2}\,(k^2 - 1) \;=\; 0.
$$
Factor out common terms where useful, then solve for $\,x\,$:
$$
2\sqrt{2}\,k\,x \;=\; -4\sqrt{2}\,(k^2 - 1)
\quad\Longrightarrow\quad
x \;=\; \frac{-4\sqrt{2}(k^2 - 1)}{2\sqrt{2}\,k}
\;=\; \frac{2(1 - k^2)}{k}.
$$
Step 4: Find y in terms of k
Substitute $\,x = \frac{2(1 - k^2)}{k}\,$ back into $\,y = \sqrt{2}x + 4\sqrt{2}\,k\,$:
$$
y \;=\; \sqrt{2} \cdot \frac{2(1 - k^2)}{k} \;+\; 4\sqrt{2}\,k
\;=\; \frac{2\sqrt{2}(1 - k^2)}{k} + 4\sqrt{2}\,k.
$$
Combine terms carefully:
$$
y \;=\; 2\sqrt{2} \,\frac{1 - k^2}{k} + 4\sqrt{2}\,k
\;=\; 2\sqrt{2} \Bigl(\frac{1 - k^2}{k} + 2\,k\Bigr)
\;=\; 2\sqrt{2} \Bigl(\frac{1 - k^2 + 2k^2}{k}\Bigr)
\;=\; 2\sqrt{2} \Bigl(\frac{1 + k^2}{k}\Bigr)
\;=\; \frac{2\sqrt{2}(1 + k^2)}{k}.
$$
Step 5: Eliminate k to find the locus in x and y
We have
$$
x \;=\; \frac{2(1 - k^2)}{k},
\quad
y \;=\; \frac{2\sqrt{2}(1 + k^2)}{k}.
$$
Rearrange each in a way that helps us form a standard conic equation.
Observe that
$$
\frac{x}{4} = \frac{2(1 - k^2)}{k} \cdot \frac{1}{4}
= \frac{1 - k^2}{2k},
\quad
\frac{y}{4\sqrt{2}} = \frac{2\sqrt{2}(1 + k^2)}{k} \cdot \frac{1}{4\sqrt{2}}
= \frac{1 + k^2}{2k}.
$$
Notice that:
$$
\Bigl(\frac{y}{4\sqrt{2}}\Bigr)^2 \;-\;\Bigl(\frac{x}{4}\Bigr)^2
= \Bigl(\frac{1 + k^2}{2k}\Bigr)^2 \;-\;\Bigl(\frac{1 - k^2}{2k}\Bigr)^2.
$$
Compute the difference of squares:
$$
\Bigl(\frac{1 + k^2}{2k}\Bigr)^2 \;-\;\Bigl(\frac{1 - k^2}{2k}\Bigr)^2
= \frac{(1 + k^2)^2 - (1 - k^2)^2}{(2k)^2}.
$$
Recall
$$
(a+b)^2 - (a-b)^2 = 4ab.
$$
Here, $\,a = 1, b = k^2\,$, so
$$
(1 + k^2)^2 - (1 - k^2)^2
= 4 \cdot 1 \cdot k^2 = 4k^2.
$$
Thus,
$$
\Bigl(\frac{1 + k^2}{2k}\Bigr)^2 \;-\;\Bigl(\frac{1 - k^2}{2k}\Bigr)^2
= \frac{4k^2}{4k^2} = 1.
$$
Therefore,
$$
\Bigl(\frac{y}{4\sqrt{2}}\Bigr)^2 \;-\;\Bigl(\frac{x}{4}\Bigr)^2
= 1.
$$
Step 6: Identify the conic and its parameters
The equation
$$
\Bigl(\frac{y}{4\sqrt{2}}\Bigr)^2 \;-\;\Bigl(\frac{x}{4}\Bigr)^2
= 1
$$
is of the standard form of a hyperbola given by:
$$
\frac{y^2}{b^2} \;-\; \frac{x^2}{a^2} \;=\; 1
\quad\Longrightarrow\quad
b^2 = (4\sqrt{2})^2 = 32,\;\; a^2 = 16.
$$
Hence, it is a hyperbola.
Step 7: Find the length of the transverse axis
For the hyperbola in this form, the transverse axis is along the $y$-direction, and its length is $2b$.
Since $\,b = 4\sqrt{2}\,$,
$$
2b = 2 \times 4\sqrt{2} = 8\sqrt{2}.
$$
Thus, the transverse axis of the hyperbola has length $\,8\sqrt{2}.$
Conclusion
The locus is a hyperbola with its transverse axis measuring $\,8\sqrt{2}\,$ in length.
