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: Recognize that the constraint $\frac{m}{s} + \frac{n}{t} = 1$ with $m+n=3$ defines an ellipse in the $(s,t)$ plane. Use Cauchy-Schwarz or the geometric property that $s+t$ is minimized when the ellipse is tangent to a line.
<p><strong>Step 1:</strong> Given $m+n=3$ (constants) and $\frac{m}{s} + \frac{n}{t} = 1$ with $s,t > 0$. We want to minimize $s+t$.</p><p><strong>Step 2:</strong> By Cauchy-Schwarz inequality: $(s+t)\left(\frac{m}{s}+\frac{n}{t}\right) \geq (\sqrt{m}+\sqrt{n})^2$</p><p><strong>Step 3:</strong> Since $\frac{m}{s}+\frac{n}{t}=1$, we have $s+t \geq (\sqrt{m}+\sqrt{n})^2 = m+n+2\sqrt{mn} = 3+2\sqrt{mn}$</p><p><strong>Step 4:</strong> Equality holds when $\frac{s}{\sqrt{m}} = \frac{t}{\sqrt{n}}$ (Cauchy-Schwarz condition), giving $s = \sqrt{m}(\sqrt{m}+\sqrt{n})$ and $t = \sqrt{n}(\sqrt{m}+\sqrt{n})$.</p><p><strong>Step 5:</strong> The minimum value of $s+t$ is $(\sqrt{m}+\sqrt{n})^2 = 3+2\sqrt{mn}$, which depends on the product $mn$. Since $m<n$ and $m+n=3$, the minimum occurs at specific values of $m,n$.</p><p>∴ Answer: C</p>
Correct Answer: C