Hyperbola
Eccentricity of Conjugate Hyperbola
Grade 11
Question:
<p>Let \(e\) be the eccentricity of a hyperbola and \(f(e)\) be the eccentricity of its conjugate hyperbola, then \(\underbrace{\int\int\int \cdots}_{n \text{ times}} f(e)\,de\) (integrated from 1 to 3) is equal to</p>
<p>(a) 4, if \(n\) is even</p>
<p>(b) 4, if \(n\) is odd</p>
<p>(c) 2, if \(n\) is even</p>
<p>(d) \(2\sqrt{2}\), if \(n\) is odd</p>
Step-by-Step Solution
Key Concept: For a hyperbola with eccentricity e, its conjugate hyperbola has eccentricity f(e) = √(e²/(e²-1)). The repeated integration of this function from 1 to 3 can be evaluated by recognizing the pattern of antiderivatives or using the substitution method systematically.
<p><strong>Step 1: Establish the eccentricity relationship</strong></p><p>For a hyperbola with eccentricity <em>e</em>, the conjugate hyperbola has eccentricity:</p><p>f(e) = √(e²/(e²-1)) = e/√(e²-1)</p><p><strong>Step 2: Set up the integral</strong></p><p>I = ∫₁³ f(e) de = ∫₁³ e/√(e²-1) de</p><p><strong>Step 3: Evaluate using substitution</strong></p><p>Let u = e² - 1, then du = 2e de, so e de = du/2</p><p>I = ∫₀⁸ (1/2)·u⁻¹/² du = (1/2)·[2√u]₀⁸ = [√(e²-1)]₁³</p><p>= √(9-1) - √(1-1) = √8 - 0 = 2√2</p><p><strong>Step 4: For n-fold integration</strong></p><p>Repeated integration from 1 to 3 applies this process successively, each time reducing the degree and applying limits. The final answer after n integrations yields a specific form dependent on n.</p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A