Let line $4 x + 5 y = 20$ intersect x – axis at A and the y – axis at B. A line L intersect AB and OA at points C and D respectively. The least value of $C D^{2}$ for which line L divides the area of $Δ O A B$ into two regions of equal area is $a \sqrt{41} - b$ where $a, b \in N$ , then $\frac{b}{a}$ is equal to

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answer is 5.

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Detailed Solution

$C (\frac{20 - 5 t}{4}, t) D (s, 0) Δ = \frac{1}{2} (5 - s) t = 5 C D^{2} = {(\frac{20 - 5 t}{4} - s)}^{2} + t^{2} = {[(\frac{10}{t}) - \frac{5 t}{4}]}^{2} + t^{2} = \frac{100}{t^{2}} + \frac{41}{16} t^{2} - 25 \geq 5 \sqrt{41} - 25$

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