<p>The minimum area of a triangle formed by any tangent to the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{81} = 1\) and the co-ordinate axes is</p>
Step-by-Step Solution
Key Concept: The tangent to an ellipse at point (x₀, y₀) is given by (xx₀/a²) + (yy₀/b²) = 1. The intercepts on the axes form a triangle whose area depends on the choice of tangent point; find the minimum by calculus.
<p><strong>Step 1:</strong> For ellipse x²/16 + y²/81 = 1, we have a = 4, b = 9.</p><p><strong>Step 2:</strong> Equation of tangent at point (4cosθ, 9sinθ) is: (xcosθ)/4 + (ysinθ)/9 = 1</p><p><strong>Step 3:</strong> x-intercept: A = 4/cosθ, y-intercept: B = 9/sinθ</p><p><strong>Step 4:</strong> Area of triangle = ½|OA||OB| = ½ · (4/cosθ) · (9/sinθ) = 18/(cosθ·sinθ)</p><p><strong>Step 5:</strong> Using cosθ·sinθ ≤ ½, we get cosθ·sinθ is maximum when θ = π/4, giving cosθ·sinθ = ½</p><p><strong>Step 6:</strong> Minimum area = 18/(½) = <strong>36</strong></p><p>∴ Answer: D</p>
Correct Answer: D