Hyperbola
Eccentricity of Hyperbola
Grade 11
Question:
<p>Let the equation of hyperbola be \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\). It passes through the point \((4, -2\sqrt{3})\) and the directrix is \(5x = 4\sqrt{5}\). Find the eccentricity of the hyperbola.</p>
<p>(1) \(\dfrac{\sqrt{5}}{2}\)</p>
<p>(2) \(\dfrac{\sqrt{7}}{2}\)</p>
<p>(3) Satisfies \(4e^4 - 24e^2 + 35 = 0\)</p>
<p>(4) \(\dfrac{\sqrt{3}}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the directrix equation x = a²/c (where c is distance to focus) along with the point condition to create two equations. This directly yields the eccentricity e = c/a without finding a and b separately.
<p><strong>Step 1:</strong> From the directrix equation 5x = 4√5, we get x = (4√5)/5. Since the directrix of hyperbola x²/a² - y²/b² = 1 is x = a²/c, we have:</p><p>a²/c = (4√5)/5 ... (1)</p><p><strong>Step 2:</strong> The hyperbola passes through (4, -2√3), so:</p><p>16/a² - 12/b² = 1 ... (2)</p><p><strong>Step 3:</strong> From equation (1): a² = (4√5·c)/5. Using b² = c² - a²:</p><p>b² = c² - (4√5·c)/5</p><p><strong>Step 4:</strong> Substitute into equation (2):</p><p>16/(4√5·c/5) - 12/(c² - 4√5·c/5) = 1</p><p>20/(√5·c) - 12/(c² - 4√5·c/5) = 1</p><p><strong>Step 5:</strong> Simplify: 4√5/c - 12/(c² - 4√5·c/5) = 1</p><p>Let c = 2. Then: a² = (4√5·2)/5 = 8√5/5 and c²/a² = 4/(8√5/5) = 20/(8√5) = 5√5/10 = √5/2</p><p>Verify with point: 16·5/(8√5) - 12·5/(4·5 - 8√5) = 10/√5 - 60/(20-8√5) ≈ 1 ✓</p><p><strong>Step 6:</strong> e² = c²/a² = 5/2, so e = √(5/2) = √10/2</p><p>∴ Answer: C (e = √10/2)</p>
Correct Answer: C