Ellipse
Auxiliary Circle and Eccentricity
Grade 11
Question:
<p>Let \(P(a\cos\theta_1, b\sin\theta_1)\), \(Q(a\cos\theta_2, b\sin\theta_2)\) and \(R(a\cos\theta_3, b\sin\theta_3)\) be the vertices of a triangle inscribed in the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). If \(\Delta_1\) = Area of \(\triangle PQR\) and \(\Delta_2\) = Area of \(\triangle P'Q'R'\) where \(P', Q', R'\) are the corresponding points on the auxiliary circle, then \(\frac{\Delta_1}{\Delta_2} = \frac{1}{7}\). Find the eccentricity of the ellipse.</p>
Step-by-Step Solution
Key Concept: The area of a triangle inscribed in an ellipse scales by factor b/a compared to the corresponding triangle on the auxiliary circle (since y-coordinates scale by b/a while x-coordinates remain unchanged). Use this ratio directly: Δ₁/Δ₂ = b/a.
<p><strong>Step 1:</strong> Set up the area ratio using parametric coordinates.</p><p>For triangle PQR inscribed in ellipse with vertices at parameters θ₁, θ₂, θ₃:</p><p>Area of △PQR = Δ₁ = ½|ab(cos θ₁(sin θ₂ - sin θ₃) + cos θ₂(sin θ₃ - sin θ₁) + cos θ₃(sin θ₁ - sin θ₂))|</p><p><strong>Step 2:</strong> For the auxiliary circle (radius a), corresponding points P', Q', R' have coordinates (a cos θᵢ, a sin θᵢ):</p><p>Area of △P'Q'R' = Δ₂ = ½|a²(cos θ₁(sin θ₂ - sin θ₃) + cos θ₂(sin θ₃ - sin θ₁) + cos θ₃(sin θ₁ - sin θ₂))|</p><p><strong>Step 3:</strong> Taking the ratio:</p><p>$$\frac{\Delta_1}{\Delta_2} = \frac{ab}{a^2} = \frac{b}{a} = \frac{1}{7}$$</p><p><strong>Step 4:</strong> Use the eccentricity relation e² = 1 - b²/a²:</p><p>$$e^2 = 1 - \left(\frac{b}{a}\right)^2 = 1 - \frac{1}{49} = \frac{48}{49}$$</p><p>$$e = \sqrt{1 - \frac{1}{49}} = \frac{4\sqrt{3}}{7}$$</p><p>∴ Answer: $\frac{b}{a} = \frac{1}{7}$, $e = \sqrt{1 - \frac{1}{49}} = \frac{4\sqrt{3}}{7}$</p>
Correct Answer: \(\frac{b}{a} = \frac{1}{7}\), \(e = \sqrt{1 - \frac{1}{49}}\)