Straight Lines
Locus
Grade 11
Question:
<p>Locus of centroid of the triangle whose vertices are \((a\cos t, a\sin t)\), \((b\sin t, -b\cos t)\) and \((1, 0)\), where <i>t</i> is a parameter, is</p>
<p>\((3x - 1)^2 + (3y)^2 = a^2 - b^2\)</p>
<p>\((3x - 1)^2 + (3y)^2 = a^2 + b^2\)</p>
<p>\((3x + 1)^2 + (3y)^2 = a^2 + b^2\)</p>
<p>\((3x + 1)^2 + (3y)^2 = a^2 - b^2\)</p>
Step-by-Step Solution
Key Concept: The centroid of a triangle is the average of its three vertices' coordinates. To find the locus, express the centroid coordinates in terms of parameter t, then eliminate t by using trigonometric identities like cos²t + sin²t = 1.
<p><strong>Step 1:</strong> Find the centroid (h, k) using the formula: centroid = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)</p><p>h = (a cos t + b sin t + 1)/3</p><p>k = (a sin t - b cos t + 0)/3</p><p><strong>Step 2:</strong> Rearrange to isolate trigonometric terms:</p><p>3h - 1 = a cos t + b sin t ... (i)</p><p>3k = a sin t - b cos t ... (ii)</p><p><strong>Step 3:</strong> Square both equations and add them:</p><p>(3h - 1)² + (3k)² = (a cos t + b sin t)² + (a sin t - b cos t)²</p><p><strong>Step 4:</strong> Expand right side:</p><p>(3h - 1)² + 9k² = a² cos²t + b² sin²t + 2ab sin t cos t + a² sin²t + b² cos²t - 2ab sin t cos t</p><p>(3h - 1)² + 9k² = a²(cos²t + sin²t) + b²(sin²t + cos²t)</p><p>(3h - 1)² + 9k² = a² + b²</p><p><strong>Step 5:</strong> Replace h with x and k with y:</p><p>(3x - 1)² + 9y² = a² + b²</p><p>∴ The locus is an ellipse: <strong>(3x - 1)² + 9y² = a² + b²</strong> or equivalently <strong>9(x - 1/3)² + 9y² = a² + b²</strong></p>
Correct Answer: B