Let the line x + y = 1 meet the axes of x and y at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is 4/9 of the area of the triangle OAB and AN : NB = $λ$ : 1, then the sum of all possible value(s) of is $λ$ :

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Detailed Solution

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Area of $ΔAOB = \frac{1}{2}$
Area of $ΔAMN = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$
Equation of AB is x + y = 1
OA = 1, AM = sec(45° – $θ$ )
AN = sec(45° – $θ$ ) cos $θ$
MN = sec(45° – $θ$ ) sin $θ$
$\begin{array}{l} Ar (△ AMN) = \frac{1}{2} \times \sec^{2} (45^{\circ} - θ) \sin θ \cdot \cos θ = \frac{2}{9} \\ \Rightarrow \tan θ = 2, \frac{1}{2} \\ \tan θ = 2 is rejected \\ \frac{AN}{NB} = \frac{λ}{1} = \cot θ = 2 \end{array}$