Hyperbola
Foci and Directrix
Grade 11
Question:
<p>Let \(a\) and \(b\), respectively, be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation \(9e^2 - 18e + 5 = 0\). If \(S(5, 0)\) is a focus and \(5x = 9\) is the corresponding directrix of this hyperbola, then \(a^2 - b^2\) is equal to</p>
<p>\(-7\)</p>
<p>\(-5\)</p>
<p>\(5\)</p>
<p>\(7\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola, use e² = 1 + b²/a² and the focus-directrix property: distance from focus to any point on hyperbola divided by distance from that point to directrix equals eccentricity. This gives ae/e = a (focus-directrix relationship) and allows finding a from the directrix equation.
<p><strong>Step 1:</strong> Solve for eccentricity using 9e² - 18e + 5 = 0</p><p>Using quadratic formula: 9e² - 18e + 5 = (3e - 1)(3e - 5) = 0</p><p>So e = 1/3 or e = 5/3. Since e > 1 for hyperbola, <strong>e = 5/3</strong></p><p><strong>Step 2:</strong> Use focus-directrix property. The focus S(5, 0) corresponds to directrix 5x = 9, or x = 9/5</p><p>For a hyperbola with center at origin and transverse axis along x-axis:</p><p>• Focus is at (ae, 0) = (5, 0) ⟹ ae = 5</p><p>• Directrix is at x = a/e = 9/5</p><p><strong>Step 3:</strong> From a/e = 9/5 and ae = 5:</p><p>Multiply: a · e · (a/e) = 5 · (9/5)</p><p>a² = 9</p><p><strong>Step 4:</strong> From ae = 5 with a = 3:</p><p>3e = 5 ⟹ e = 5/3 ✓ (confirms our eccentricity)</p><p><strong>Step 5:</strong> Use e² = 1 + b²/a²:</p><p>(5/3)² = 1 + b²/9</p><p>25/9 = 1 + b²/9</p><p>b²/9 = 25/9 - 9/9 = 16/9</p><p>b² = 16</p><p><strong>Step 6:</strong> Therefore a² - b² = 9 - 16 = -7</p><p>∴ Answer: <strong>-7</strong> (Option A)</p>
Correct Answer: A