<p>\(A = (\sqrt{1-t^2} + t, 0)\) and \(B = (\sqrt{1-t^2} - t, 2t)\) are two variable points where \(t\) is a parameter. The locus of the middle point of AB is</p>
Step-by-Step Solution
Key Concept: Find the midpoint coordinates in terms of parameter t, then eliminate t to get the locus equation. The constraint from the parametrization √(1-t²) being real forces |t| ≤ 1, which translates to a domain restriction on the locus.
<p><strong>Step 1:</strong> Find midpoint M of AB.</p><p>M = ((√(1-t²) + t + √(1-t²) - t)/2, (0 + 2t)/2)</p><p>M = (√(1-t²), t)</p><p><strong>Step 2:</strong> Let the midpoint be (x, y).</p><p>x = √(1-t²) and y = t</p><p><strong>Step 3:</strong> Eliminate parameter t.</p><p>From y = t, we get t = y</p><p>Substituting in x = √(1-t²):</p><p>x = √(1-y²)</p><p>Squaring both sides: x² = 1 - y²</p><p>∴ x² + y² = 1</p><p><strong>Step 4:</strong> Apply domain restriction.</p><p>Since x = √(1-t²) ≥ 0 and |t| ≤ 1 (for √(1-t²) to be real),</p><p>The locus is x² + y² = 1 with x ≥ 0 and -1 ≤ y ≤ 1</p><p>This represents the <strong>right semicircle</strong> of radius 1 centered at origin.</p><p>∴ Answer: C</p>
Correct Answer: C