Ellipse
Chord with given midpoint
Grade 11
Question:
<p><strong>264.</strong> Given that \(m, n, s, t \in (0, +\infty)\), \(m + n = 3\), \(\dfrac{m}{s} + \dfrac{n}{t} = 1\), \(m, n\) are constants and \(m < n\). If the minimum value of \(s + t\) is \(3 + 2\sqrt{2}\), point \((m, n)\) is the mid-point of a chord of the ellipse \(\dfrac{x^2}{4} + \dfrac{y^2}{16} = 1\). Find the equation of the line where the chord lies:</p>
<p>(a) \(x + y - 3 = 0\)</p>
<p>(b) \(x - 2y + 3 = 0\)</p>
<p>(c) \(2x + y - 4 = 0\)</p>
<p>(d) \(4x + 2y - 3 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the constraint ∑(m/s + n/t) = 1 with Cauchy-Schwarz inequality. The minimum of s + t occurs when the weighted sum is optimized using the condition that variables are positive and m + n = 3 is fixed.
<p><strong>Step 1:</strong> Given constraints: m + n = 3, m/s + n/t = 1, where m, n, s, t ∈ (0, +∞)</p><p><strong>Step 2:</strong> Apply Cauchy-Schwarz inequality: (m/s + n/t)(s + t) ≥ (√m + √n)²</p><p>Since m/s + n/t = 1, we have: 1·(s + t) ≥ (√m + √n)²</p><p><strong>Step 3:</strong> Expand: s + t ≥ m + n + 2√(mn) = 3 + 2√(mn)</p><p><strong>Step 4:</strong> To minimize s + t, we need to minimize √(mn) subject to m + n = 3. By AM-GM: mn ≤ (m+n)²/4 = 9/4, with minimum approaching 0 as one variable approaches 0 (but this contradicts optimality in the ellipse context).</p><p><strong>Step 5:</strong> The equality condition in Cauchy-Schwarz requires m/s = n/t (proportionality). Combined with m/s + n/t = 1, this gives s = 2m and t = 2n.</p><p><strong>Step 6:</strong> Therefore: s + t = 2m + 2n = 2(m + n) = 2(3) = 6</p><p>∴ Answer: C (minimum value of s + t is 6)</p>
Correct Answer: C