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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Straight Lines

Let the line x+y=1 meet the axes of x and y at A and B,...

Let the line x+y=1x+y=1 meet the axes of xx and yy 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 OBO B and ABA B, respectively. If the area of the triangle AMNA M N is 49\frac{4}{9} of the area of the triangle OABO A B and AN : NB =λ:1=\lambda: 1, then the sum of all possible value(s) of is λ\lambda :

Options

  1. A2
  2. B52\frac{5}{2}
  3. C12\frac{1}{2}
  4. D136\frac{13}{6}

Correct answer

A. 2

Step-by-step solution

 Area of  △AOB=12 Area of △AMN=49×12=29\begin{aligned} & \text{ Area\text{ 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 x+y=1\mathrm{x}+\mathrm{y}=1OA=1,AM=sec⁡(45∘−θ)AN=sec⁡(45∘−θ)cos⁡θMN=sec⁡(45∘−θ)sin⁡θ\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}Ar⁡(△AMN)=12×sec⁡2(45∘−θ)sin⁡θ⋅cos⁡θ=29⇒tan⁡θ=2,12tan⁡θ=2 is rejected ANNB=λ1=cot⁡θ=2\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}$$
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