Question:
<p>The locus of the point of intersection of the straight lines, \(tx - 2y - 3t = 0\); \(x - 2ty + 3 = 0\) (\(t \in \mathbb{R}\)), is</p>
<p>a hyperbola with the length of conjugate axis 3.</p>
<p>an ellipse with eccentricity \(\dfrac{2}{\sqrt{5}}\).</p>
<p>an ellipse with the length of major axis 6.</p>
<p>a hyperbola with eccentricity \(\sqrt{5}\).</p>
Step-by-Step Solution
Key Concept: Rewrite both lines by grouping terms with parameter t separately, then eliminate t by using the condition that both equations must be satisfied simultaneously for points on the locus.
<p><strong>Step 1:</strong> Rewrite the given lines by factoring out parameter t:</p><p>Line 1: tx - 2y - 3t = 0 → t(x - 3) = 2y</p><p>Line 2: x - 2ty + 3 = 0 → x + 3 = 2ty</p><p><strong>Step 2:</strong> From Line 1: t = 2y/(x-3) (provided x ≠ 3)</p><p>From Line 2: t = (x+3)/(2y) (provided y ≠ 0)</p><p><strong>Step 3:</strong> Equate the two expressions for t:</p><p>2y/(x-3) = (x+3)/(2y)</p><p><strong>Step 4:</strong> Cross-multiply:</p><p>4y² = (x-3)(x+3)</p><p>4y² = x² - 9</p><p><strong>Step 5:</strong> Rearrange to standard form:</p><p>x² - 4y² = 9</p><p>x²/9 - 4y²/9 = 1</p><p>x²/9 - y²/(9/4) = 1</p><p>∴ The locus is a hyperbola: <strong>x² - 4y² = 9</strong></p>
Correct Answer: A