Hyperbola
Eccentricity of Hyperbola
Grade 11
Question:
<p>The eccentricity of the hyperbola <br> \(16x^2 - 9y^2 = 144\) is</p>
<p>(1) \(\dfrac{3}{5}\)</p>
<p>(2) \(\dfrac{4}{5}\)</p>
<p>(3) \(\dfrac{3}{4}\)</p>
<p>(4) \(\dfrac{5}{3}\)</p>
Step-by-Step Solution
Key Concept: Convert the hyperbola to standard form x²/a² - y²/b² = 1 by dividing by 144, then use e = √(1 + b²/a²) for a hyperbola where a² and b² are the denominators.
<p><strong>Step 1:</strong> Convert to standard form by dividing the equation 16x² - 9y² = 144 by 144:</p><p>16x²/144 - 9y²/144 = 1</p><p>x²/9 - y²/16 = 1</p><p><strong>Step 2:</strong> Identify a² = 9 and b² = 16, so a = 3 and b = 4</p><p><strong>Step 3:</strong> For a hyperbola, the relationship is c² = a² + b² (not c² = a² - b²)</p><p>c² = 9 + 16 = 25</p><p>c = 5</p><p><strong>Step 4:</strong> Calculate eccentricity e = c/a = 5/3</p><p>∴ Answer: e = 5/3</p>
Correct Answer: D