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 $O B$ and $A B$, respectively. If the area of the triangle $A M N$ is $\frac{4}{9}$ of the area of the triangle $O A B$ and AN : NB $=\lambda: 1$, then the sum of all possible value(s) of is $\lambda$ :

• Correct answer
  A

  2
• B

  $\frac{5}{2}$
• C

  $\frac{1}{2}$
• D

  $\frac{13}{6}$

Solution

$\begin{aligned} & \text { Area of } \triangle \mathrm{AOB}=\frac{1}{2} \\ & \text { Area of } \triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9} \end{aligned}$ Equation of AB is $\mathrm{x}+\mathrm{y}=1$ $\begin{aligned} & \mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right) \\ & \mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta \\ & \mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta \end{aligned}$ $\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{AMN})=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9} \\ & \Rightarrow \tan \theta=2, \frac{1}{2} \\ & \tan \theta=2 \text { is rejected } \\ & \frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda}{1}=\cot \theta=2\end{aligned}$

Hence, the correct answer is option: 'A'.