JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 8)
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 $ \frac{4}{9} $ of the area of the triangle OAB and AN : NB = $ \lambda : 1 $, then the sum of all possible value(s) of $ \lambda $ is:
$\frac{1}{2}$
$\frac{5}{2}$
2
$\frac{13}{6}$
Explanation
_29th_January_Evening_Shift_en_8_1.png)
Area of $\triangle \mathrm{AOB}=\frac{1}{2}$
Area of $\triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$
Equation of AB is $\mathrm{x}+\mathrm{y}=1$
$$\begin{aligned} & \mathrm{OA}=1, \mathrm{AM}=\sec \left(45^{\circ}-\theta\right) \\ & \mathrm{AN}=\sec \left(45^{\circ}-\theta\right) \cos \theta \\ & \mathrm{MN}=\sec \left(45^{\circ}-\theta\right) \sin \theta \end{aligned}$$
$$\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{AMN})=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9} \\ & \Rightarrow \tan \theta=2, \frac{1}{2} \\ & \tan \theta=2 \text { is rejected } \\ & \frac{\mathrm{AN}}{\mathrm{NB}}=\frac{\lambda}{1}=\cot \theta=2 \end{aligned}$$
Comments (0)
