<p>Let \(\Delta_1\) be the area of a triangle \(PQR\) inscribed in an ellipse and \(\Delta_2\) be the area of the triangle \(P'Q'R'\) whose vertices are the points lying on the auxiliary circle corresponding to the points \(P\), \(Q\), \(R\) respectively. If the eccentricity of the ellipse is \(\dfrac{4\sqrt{3}}{7}\) then the ratio \(\dfrac{\Delta_2}{\Delta_1}\) is equal to</p>
Step-by-Step Solution
Key Concept: When a point on an ellipse is mapped to the auxiliary circle via the eccentric angle, the area scales by a constant factor equal to b/a (the ratio of semi-minor to semi-major axis). This ratio is independent of which triangle is inscribed.
<p><strong>Step 1:</strong> Recall the parametric form: a point P on ellipse is (a cos φ, b sin φ), and the corresponding point P' on the auxiliary circle (radius a) is (a cos φ, a sin φ).</p><p><strong>Step 2:</strong> For any triangle with vertices at parameters φ₁, φ₂, φ₃:</p><p>Δ₁ = (ab/2)|sin(φ₂ - φ₁) + sin(φ₃ - φ₂) + sin(φ₁ - φ₃)|</p><p>Δ₂ = (a²/2)|sin(φ₂ - φ₁) + sin(φ₃ - φ₂) + sin(φ₁ - φ₃)|</p><p><strong>Step 3:</strong> Taking the ratio: Δ₂/Δ₁ = a/b</p><p><strong>Step 4:</strong> Given e = 4√3/7, find b/a:</p><p>e² = 1 - b²/a² ⇒ (4√3/7)² = 1 - b²/a²</p><p>48/49 = 1 - b²/a² ⇒ b²/a² = 1/49 ⇒ b/a = 1/7</p><p><strong>Step 5:</strong> Therefore: Δ₂/Δ₁ = a/b = 7</p><p>∴ <strong>Answer: 7</strong></p>
Correct Answer: 7