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…

Question

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 $\lambda$ :

Aicharya · Accepted Answer

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

Let the line $x+y=1$ meet the axes of $x$ and $y$ at A and B, respectively. A right angled triangle AMN is…

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