Question:

Step 1: Find the minimum value of $s+t$ using the constraint equations. We are given that $m+n=3$ and $\frac{m}{s}+\frac{n}{t}=1$. We need to minimize $s+t$. Multiply the constraint by $(s+t)$: $$\left(\frac{m}{s}+\frac{n}{t}\right)(s+t) = 1 \cdot (s+t)$$ Expanding the left side: $$m + \frac{mt}{s} + \frac{ns}{t} + n = s+t$$ Rearranging: $$s+t = m+n + \frac{mt}{s} + \frac{ns}{t} = 3 + \frac{mt}{s} + \frac{ns}{t}$$ Step 2: Apply AM-GM inequality to find the minimum. By the AM-GM inequality: $$\frac{mt}{s} + \frac{ns}{t} \geq 2\sqrt{\frac{mt}{s} \cdot \frac{ns}{t}} = 2\sqrt{mn}$$ Therefore: $$s+t \geq 3 + 2\sqrt{mn}$$ Step 3: Use the given minimum value to find $mn$. We are given that the minimum value of $s+t$ is $3+2\sqrt{2}$. Thus: $$3 + 2\sqrt{mn} = 3 + 2\sqrt{2}$$ This gives us: $$\sqrt{mn} = \sqrt{2} \implies mn = 2$$ Step 4: Solve for $m$ and $n$. We have the system: - $m+n=3$ - $mn=2$ These are the sum and product of roots, so $m$ and $n$ satisfy: $$x^2 - 3x + 2 = 0$$ Factoring: $$(x-1)(x-2) = 0$$ So $m=1$ and $n=2$ (since $m < n$). Step 5: Identify the midpoint of the chord. The point $(m,n) = (1,2)$ is the midpoint of a chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{16}=1$. Step 6: Find the equation of the chord using the midpoint condition. For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a^2=4$ and $b^2=16$, if a chord has midpoint $(x_0, y_0)$, the equation of the chord is given by: $$\frac{x \cdot x_0}{a^2} + \frac{y \cdot y_0}{b^2} = \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2}$$ Substituting $(x_0, y_0) = (1,2)$, $a^2=4$, and $b^2=16$: $$\frac{x \cdot 1}{4} + \frac{y \cdot 2}{16} = \frac{1^2}{4} + \frac{2^2}{16}$$ $$\frac{x}{4} + \frac{y}{8} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ Multiplying through by 4: $$x + \frac{y}{2} = 2$$ Multiplying through by 2: $$2x + y = 4$$ Step 7: State the final answer. The equation of the line where the chord lies is: $$2x + y - 4 = 0$$ The answer is **Option 3**.