Question:
<p>The locus of the point of intersection of the lines, \(\sqrt{2}x - y + 4\sqrt{2}k = 0\) and \(\sqrt{2}kx + ky - 4\sqrt{2} = 0\) (\(k\) is any non-zero real parameter), is</p>
<p>an ellipse whose eccentricity is \(\dfrac{1}{\sqrt{3}}\).</p>
<p>an ellipse with length of its major axis \(8\sqrt{2}\).</p>
<p>a hyperbola whose eccentricity is \(\sqrt{3}\).</p>
<p>a hyperbola with length of its transverse axis \(8\sqrt{2}\).</p>
Step-by-Step Solution
Key Concept: Eliminate the parameter k from two linear equations by expressing k from one equation and substituting into the other, or by finding a relationship between x and y that holds for all values of k.
<p><strong>Step 1:</strong> From the first equation: √2·x − y + 4√2·k = 0, we get y = √2·x + 4√2·k</p><p><strong>Step 2:</strong> From the second equation: √2·k·x + k·y − 4√2 = 0, factor out k: k(√2·x + y) = 4√2, so √2·x + y = 4√2/k (for k ≠ 0)</p><p><strong>Step 3:</strong> From Step 2: k = 4√2/(√2·x + y). Substitute into y = √2·x + 4√2·k: y = √2·x + 4√2 · [4√2/(√2·x + y)]</p><p><strong>Step 4:</strong> Simplify: y = √2·x + 32/(√2·x + y). Multiply by (√2·x + y): y(√2·x + y) = √2·x(√2·x + y) + 32</p><p><strong>Step 5:</strong> Expand: √2·xy + y² = 2x² + √2·xy + 32. Cancel √2·xy: y² = 2x² + 32</p><p><strong>Step 6:</strong> Rearrange: 2x² − y² + 32 = 0, or equivalently: y²/32 − x²/16 = 1</p><p>∴ <strong>Answer: D</strong> (A rectangular hyperbola with the standard form y²/32 − x²/16 = 1)</p>
Correct Answer: D