Ellipse
Maximum area of inscribed rectangle
Grade 11
Question:
<p>The equation of ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Parametric equation is \(x = a\cos\theta,\ y = b\sin\theta\). Let the coordinates of \(A\) be \((a\cos\theta,\ b\sin\theta)\); \(\theta \in (0, \pi/2)\). A rectangle is inscribed in the ellipse with sides parallel to the axes. The maximum area of the rectangle inscribed in the ellipse is:</p>
<p>\(ab\)</p>
<p>\(2ab\)</p>
<p>\(\frac{ab}{2}\)</p>
<p>\(4ab\)</p>
Step-by-Step Solution
Key Concept: Use the parametric form to express rectangle vertices and maximize area = 4xy = 4ab·cos(θ)sin(θ) = 2ab·sin(2θ), which peaks when sin(2θ) = 1.
<p><strong>Step 1:</strong> Since A = (a cos θ, b sin θ) lies on the ellipse in the first quadrant (θ ∈ (0, π/2)), the inscribed rectangle with sides parallel to axes has vertices at (±a cos θ, ±b sin θ).</p><p><strong>Step 2:</strong> The dimensions of the rectangle are:</p><ul><li>Length = 2a cos θ</li><li>Width = 2b sin θ</li></ul><p><strong>Step 3:</strong> Area A(θ) = (2a cos θ)(2b sin θ) = 4ab cos θ sin θ = 2ab sin(2θ)</p><p><strong>Step 4:</strong> To maximize: dA/dθ = 2ab · 2cos(2θ) = 4ab cos(2θ)</p><p>Setting dA/dθ = 0: cos(2θ) = 0 ⟹ 2θ = π/2 ⟹ θ = π/4</p><p><strong>Step 5:</strong> At θ = π/4: sin(2θ) = sin(π/2) = 1</p><p>Maximum Area = 2ab(1) = <strong>2ab</strong></p><p>∴ Answer: B</p>
Correct Answer: B