Verified Solution • 2025–26 Session moderate

QJEEM25Mabfda: JEE Main 2025 Mathematics - Class 11 JEEM25Mabfda Solution: Let the line x + y = 1 meet the axes of x and y at A and B ,…

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 49\frac{4}{9} of the area of the triangle OAB and AN : NB = λ:1\lambda : 1 , then the sum of all possible value(s) of λ\lambda is:

Multiple Choice Options

  • A 12\frac{1}{2}
  • B 52\frac{5}{2}
  • C 2
  • D 136\frac{13}{6}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 29th January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 8 English Explanation

Area of △AOB=12\triangle \mathrm{AOB}=\frac{1}{2}

Area of △AMN=49×12=29\triangle \mathrm{AMN}=\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}

Equation of AB is x+y=1\mathrm{x}+\mathrm{y}=1

amp;OA=1,AM=sec⁡(45∘−θ)amp;AN=sec⁡(45∘−θ)cos⁡θamp;MN=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}

amp;Ar⁡(△AMN)=12×sec⁡2(45∘−θ)sin⁡θ⋅cos⁡θ=29amp;⇒tan⁡θ=2,12amp;tan⁡θ=2 is rejected amp;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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