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MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

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 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:1AN : NB = \lambda : 1, then the sum of all possible values of λ\lambda is:

  • A

    12\frac{1}{2}

  • B

    136\frac{13}{6}

  • C

    55

  • D

    22

Answer

Correct answer:B

Step-by-step solution

Coordinate Geometry Method

Given: The line x+y=1x + y = 1 meets the coordinate axes at A and B. Triangle AMN is right-angled, with M on OB and N on AB. Also, Area(△AMN)=49 Area(△OAB)\text{Area}(\triangle AMN)=\frac{4}{9}\,\text{Area}(\triangle OAB) and AN:NB=λ:1AN:NB=\lambda:1.

Find: The sum of all possible values of λ\lambda.

First find the intercepts of the line x+y=1x + y = 1:

x=1⇒A(1,0)x=1 \Rightarrow A(1,0) y=1⇒B(0,1)y=1 \Rightarrow B(0,1)

Hence, O=(0,0)O=(0,0), A=(1,0)A=(1,0) and B=(0,1)B=(0,1).

The area of triangle OABOAB is

Area(△OAB)=12×1×1=12\text{Area}(\triangle OAB)=\frac{1}{2}\times 1 \times 1=\frac{1}{2}

Therefore,

Area(△AMN)=49×12=29\text{Area}(\triangle AMN)=\frac{4}{9}\times \frac{1}{2}=\frac{2}{9}

Let M=(0,m)M=(0,m) on OB. Since AN:NB=λ:1AN:NB=\lambda:1, point NN divides AB internally in the ratio λ:1\lambda:1. So by section formula,

N=(λλ+1,1λ+1)N=\left(\frac{\lambda}{\lambda+1},\frac{1}{\lambda+1}\right)

Using the coordinate area formula for triangle with vertices A(1,0)A(1,0), M(0,m)M(0,m) and N(λλ+1,1λ+1)N\left(\frac{\lambda}{\lambda+1},\frac{1}{\lambda+1}\right),

Area=12∣1(m−1λ+1)+0+λλ+1(0−m)∣\text{Area}=\frac{1}{2}\left|1\left(m-\frac{1}{\lambda+1}\right)+0+\frac{\lambda}{\lambda+1}(0-m)\right| =12∣m−1λ+1−mλλ+1∣=\frac{1}{2}\left|m-\frac{1}{\lambda+1}-\frac{m\lambda}{\lambda+1}\right| =12∣m−1λ+1∣=\frac{1}{2}\left|\frac{m-1}{\lambda+1}\right|

Equating this to 29\frac{2}{9},

12⋅∣m−1∣λ+1=29\frac{1}{2}\cdot \frac{|m-1|}{\lambda+1}=\frac{2}{9} ∣m−1∣=49(λ+1)|m-1|=\frac{4}{9}(\lambda+1)

Since m<1m<1, we take

m=1−49(λ+1)=5−4λ9m=1-\frac{4}{9}(\lambda+1)=\frac{5-4\lambda}{9}

Now use the right angle condition at M:

AM⊥MNAM \perp MN

Slope of AMAM is

m−00−1=−m\frac{m-0}{0-1}=-m

Slope of MNMN is

m−1λ+10−λλ+1=−m(λ+1)−1λ\frac{m-\frac{1}{\lambda+1}}{0-\frac{\lambda}{\lambda+1}}=-\frac{m(\lambda+1)-1}{\lambda}

Therefore,

(−m)(−m(λ+1)−1λ)=−1(-m)\left(-\frac{m(\lambda+1)-1}{\lambda}\right)=-1 m[m(λ+1)−1]λ=−1\frac{m[m(\lambda+1)-1]}{\lambda}=-1 m2(λ+1)−m+λ=0m^2(\lambda+1)-m+\lambda=0

Substitute m=5−4λ9m=\frac{5-4\lambda}{9}:

(5−4λ9)2(λ+1)−5−4λ9+λ=0\left(\frac{5-4\lambda}{9}\right)^2(\lambda+1)-\frac{5-4\lambda}{9}+\lambda=0

Multiplying through by 8181,

(5−4λ)2(λ+1)−9(5−4λ)+81λ=0(5-4\lambda)^2(\lambda+1)-9(5-4\lambda)+81\lambda=0 (25−40λ+16λ2)(λ+1)−45+36λ+81λ=0(25-40\lambda+16\lambda^2)(\lambda+1)-45+36\lambda+81\lambda=0 16λ3−24λ2+102λ−20=016\lambda^3-24\lambda^2+102\lambda-20=0 8λ3−12λ2+51λ−10=08\lambda^3-12\lambda^2+51\lambda-10=0

Solving this cubic gives the valid values

λ=12,λ=103\lambda=\frac{1}{2},\quad \lambda=\frac{10}{3}

Hence, the sum of all possible values is

λ1+λ2=12+103=136\lambda_1+\lambda_2=\frac{1}{2}+\frac{10}{3}=\frac{13}{6}

Therefore, the correct option is B.

Area and Ratio-Based Approach

Given: Area(△OAB)=12\text{Area}(\triangle OAB)=\frac{1}{2} because the intercepts are each 11, and Area(△AMN)=29\text{Area}(\triangle AMN)=\frac{2}{9}.

Find: The possible values of λ\lambda and their sum.

From the alternate approach in the solution, let the right triangle geometry be expressed using an angle θ\theta. Then

AM=sec⁡(45∘−θ)AM=\sec(45^\circ-\theta) AN=sec⁡(45∘−θ)cos⁡θ,MN=sec⁡(45∘−θ)sin⁡θAN=\sec(45^\circ-\theta)\cos\theta, \quad MN=\sec(45^\circ-\theta)\sin\theta

So,

Area(△AMN)=12sec⁡2(45∘−θ)sin⁡θcos⁡θ\text{Area}(\triangle AMN)=\frac{1}{2}\sec^2(45^\circ-\theta)\sin\theta\cos\theta

Given this area equals 29\frac{2}{9}, the working yields

tan⁡θ=2or12\tan\theta=2 \quad \text{or} \quad \frac{1}{2}

The value tan⁡θ=2\tan\theta=2 is rejected, so

tan⁡θ=12\tan\theta=\frac{1}{2}

Using the similarity condition stated in the solution,

ANNB=λ\frac{AN}{NB}=\lambda

which gives

λ=136\lambda=\frac{13}{6}

Thus the required sum of all possible values of λ\lambda is 136\frac{13}{6}.

Common mistakes

  • Assuming NN is the midpoint of AB. This is wrong because the question gives the variable ratio AN:NB=λ:1AN:NB=\lambda:1. Use the section formula with the given ratio instead.

  • Using the wrong right-angle location. The perpendicular condition must be applied at M, so slope(AM)×slope(MN)=−1\text{slope}(AM)\times \text{slope}(MN)=-1. Applying it at another vertex leads to an incorrect equation in λ\lambda.

  • Computing the area of triangle AMN incorrectly by taking a convenient base and height without checking perpendicularity. Use the coordinate area formula directly unless the chosen base-height pair is certainly perpendicular.

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