Ellipse
Latus Rectum and Eccentricity
Grade 11
Question:
<p>Let <span>\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)</span> <span>(a > b)</span> be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function <span>\(f(t) = \frac{5}{12} + t - t^2\)</span>, then <span>\(a^2 + b^2\)</span> is equal to</p>
<p>(a) 145</p>
<p>(b) 116</p>
<p>(c) 126</p>
<p>(d) 135</p>
Step-by-Step Solution
Key Concept: Use the condition on latus rectum to relate a and b, find the maximum of the given function to determine eccentricity, then solve for a² + b².
<p><strong>Step 1:</strong> Length of latus rectum = <span>\(\frac{2b^2}{a} = 10\)</span> [given]</p><p>Therefore, <span>\(b^2 = 5a\)</span> ... (i)</p><p><strong>Step 2:</strong> Find maximum of <span>\(f(t) = \frac{5}{12} + t - t^2\)</span></p><p>Taking derivative: <span>\(f'(t) = 1 - 2t = 0\)</span></p><p>Thus <span>\(t = \frac{1}{2}\)</span></p><p><strong>Step 3:</strong> Maximum value is <span>\(f\left(\frac{1}{2}\right) = \frac{5}{12} + \frac{1}{2} - \frac{1}{4} = \frac{5 + 6 - 3}{12} = \frac{2}{3}\)</span></p><p><strong>Step 4:</strong> Eccentricity <span>\(e = \frac{2}{3}\)</span></p><p>We know <span>\(e^2 = 1 - \frac{b^2}{a^2}\)</span></p><p><span>\(\frac{4}{9} = 1 - \frac{b^2}{a^2}\)</span></p><p><span>\(\frac{b^2}{a^2} = \frac{5}{9}\)</span></p><p>From equation (i): <span>\(b^2 = 5a\)</span>, so <span>\(\frac{5a}{a^2} = \frac{5}{9}\)</span></p><p><span>\(a = 9\)</span> and <span>\(b^2 = 45\)</span></p><p><strong>Step 5:</strong> <span>\(a^2 + b^2 = 81 + 45 = 126\)</span></p><p>∴ Answer is (c).</p>
Correct Answer: c