Ellipse
Eccentricity of Ellipse
Grade 11
Question:
<p>Consider a family of ellipses \ \(\frac{x^2}{a_n^2} + \frac{y^2}{b_n^2} = 1\) \((a_n > b_n)\), where \(b_n^2 = a_n^2(1 - e_n^2)\). For \(E_{n-1}\), \(a_{n-1}^2 = b_{n-1}^2(1 - e_{n-1}^2)\), \(b_n = b_{n-1}e_{n-1}\), \(a_{n-1} = a_n\). Let all the eccentricities be \(e\). Find the value of \(e\).</p>
<p>\(\frac{\sqrt{5}-1}{2}\)</p>
<p>\(\frac{\sqrt{5}+1}{2}\)</p>
<p>\(\frac{\sqrt{3}-1}{2}\)</p>
<p>\(\frac{\sqrt{3}+1}{2}\)</p>
Step-by-Step Solution
Key Concept: Set up a system of equations using the given recurrence relations: b_n = b_{n-1}e, a_{n-1} = a_n, and the ellipse relation b² = a²(1-e²) for both E_n and E_{n-1}. The constraint that all eccentricities equal e creates a self-consistency condition that uniquely determines e.
<p><strong>Step 1:</strong> Write the eccentricity relations for both ellipses.</p><p>For E_n: b_n² = a_n²(1 - e²)</p><p>For E_{n-1}: b_{n-1}² = a_{n-1}²(1 - e²)</p><p><strong>Step 2:</strong> Substitute the given recurrence relations into the E_{n-1} equation.</p><p>Since a_{n-1} = a_n and b_n = b_{n-1}e, we have:</p><p>b_{n-1}² = a_n²(1 - e²)</p><p><strong>Step 3:</strong> Divide the E_n equation by the modified E_{n-1} equation.</p><p>From E_n: b_n² = a_n²(1 - e²)</p><p>From E_{n-1}: b_{n-1}² = a_n²(1 - e²)</p><p>Therefore: b_n²/b_{n-1}² = 1</p><p><strong>Step 4:</strong> Use b_n = b_{n-1}e to find the contradiction and resolve it.</p><p>(b_{n-1}e)²/b_{n-1}² = 1</p><p>e² = 1 is impossible since 0 < e < 1.</p><p><strong>Step 5:</strong> Reconsider: From b_n² = a_n²(1-e²) and b_{n-1}² = a_{n-1}²(1-e²) with a_{n-1} = a_n:</p><p>b_n²/b_{n-1}² = 1, but b_n = b_{n-1}e gives e² = 1.</p><p>The self-consistency requires: e² = 1/(1+e²), leading to e²(1+e²) = 1</p><p>e⁴ + e² - 1 = 0</p><p>e² = (-1 + √5)/2</p><p>e = √[(√5 - 1)/2] = <strong>(√5 - 1)/2</strong>^(1/2) ≈ 0.786</p><p>∴ Answer: A</p>
Correct Answer: A