Given that $m, n, s, t \in (0, \infty); m + n = 3; \frac{m}{s} + \frac{n}{t} = 1; m, n$ are Constants and $m < n$ . If the minimum value of $s + t$ is $3 + 2 \sqrt{2}$ , point $(m, n)$ is the midpoint of a chord of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{16} = 1$ the equation of the line where the chord lies

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$x + y - 3 = 0$

$x - 2 y + 3 = 0$

$4 x + 2 y - 3 = 0$

$2 x + y - 4 = 0$

answer is C.

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Detailed Solution

$\frac{m}{s} = \sin^{2} θ; \frac{n}{t} = \cos^{2} θ \Rightarrow s = m {cosec}^{2} θ t = n \sec^{2} θ$

$s + t = m {cosec}^{2} θ + n \sec^{2} θ$

$= m + n + m \cot^{2} θ + n \tan^{2} θ$

$= 3 + \underset{U s e A m ⩾ G m}{\underset{⏟}{m {cot}^{2} θ + n + n + n \tan^{2} θ}}$

$\geq 3 + 2 \sqrt{m n}$ $∴ m n = 2$ $m + n = 3$

$∴ m = 1$ $n = 2 (∴ m < n)$ Point is $(1, 2)$

uses $S_{1} = S_{11}$

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