Conic section and inequality

Combining $m + n = 3$ and $\frac{m}{s} + \frac{n}{t} = 1$ , and solving for $s$ gives $s = \frac{mt}{t+m-3}$ . If $y = s + t = \frac{mt}{t + m - 3} + t$ , then the minimum value is when the derivative $y' = 0$ , or when $y' = \frac{m(t + m - 3) - mt}{(t + m - 3)^2} + 1 = 0$ . Solving this for $t$ gives $t = -m + 3 + \sqrt{-m^2 + 3m}$ , and then solving for $s + t$ gives $s + t = 3 + 2\sqrt{-m^2 + 3m}$ . Since the minimum value of $s + t$ is $3 + 2\sqrt{2}$ , $-m^2 + 3m = 2$ , which solves to $m = 1$ or $m = 2$ . If $m = 2$ then $n = 1$ , but $m < n$ . Therefore, $(m, n) = (1, 2)$ .

The ellipse $\frac{x^2}{4} + \frac{y^2}{16} = 1$ can be converted to a circle with a radius of $2$ if all the $y$ -values are multiplied by $\frac{1}{2}$ . Applying this to the point $(1, 2)$ gives a new transformed point of $(1, 1)$ . Since the slope between $(0, 0)$ and $(1, 1)$ is $1$ , a chord perpendicular to this and through $(1, 1)$ would have an equation of $y = -x + 2$ and pass through $(0, 2)$ . Converting back to the ellipse (by multiplying all the $y$ -values by $2$ ) would give a chord of an ellipse that passes through $(1, 2)$ and $(0, 4)$ , which is the line $\boxed{2x + y - 4 = 0}$ .