<p>The locus of the foot of perpendicular drawn from the centre of the ellipse \(x^2 + 3y^2 = 6\) on any tangent to it is</p>
<p>\((x^2 + y^2)^2 = 6x^2 + 2y^2\)</p>
<p>\((x^2 + y^2)^2 = 6x^2 - 2y^2\)</p>
<p>\((x^2 - y^2)^2 = 6x^2 + 2y^2\)</p>
<p>\((x^2 - y^2)^2 = 6x^2 - 2y^2\)</p>
Step-by-Step Solution
Key Concept: The locus of the foot of perpendicular from the center of an ellipse to any tangent forms a curve whose equation can be derived using the parametric form of the tangent and the perpendicularity condition. For ellipse x²/a² + y²/b² = 1, this locus is the auxiliary circle x² + y² = a² + b².
<p><strong>Step 1:</strong> Convert the ellipse to standard form: x²/6 + y²/2 = 1</p><p>Here a² = 6, b² = 2, so center is at origin (0,0).</p><p><strong>Step 2:</strong> The parametric form of tangent to the ellipse is: (x/√6)cos θ + (y/√2)sin θ = 1</p><p><strong>Step 3:</strong> Let P(h,k) be the foot of perpendicular from O(0,0) to this tangent. Then OP is perpendicular to the tangent.</p><p><strong>Step 4:</strong> The perpendicular from origin has direction ratios proportional to the normal of tangent: (cos θ/√6, sin θ/√2)</p><p><strong>Step 5:</strong> Since P lies on the tangent and OP is perpendicular to it: P divides the perpendicular such that the distance from O to tangent is r = OP.</p><p><strong>Step 6:</strong> Distance from (0,0) to tangent (x/√6)cos θ + (y/√2)sin θ - 1 = 0 is:</p><p>r = 1/√(cos²θ/6 + sin²θ/2) = 1/√(cos²θ/6 + sin²θ/2)</p><p><strong>Step 7:</strong> For the locus, we use: h² + k² = r² = 1/(cos²θ/6 + sin²θ/2) = 6·2/(2cos²θ + 6sin²θ) = 12/(2 + 4sin²θ)</p><p><strong>Step 8:</strong> Eliminating θ using the tangent foot relationship: The locus becomes x² + y² = a² + b² = 6 + 2 = 8</p><p>∴ Answer: <strong>x² + y² = 8</strong></p>
Correct Answer: A