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 poins M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN i

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Solution

Q.
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 = $λ$ : 1, then the sum of all possible value(s) of $λ$ is : [2025]

1 $\frac{5}{2}$

2 $\frac{13}{6}$

3 $\frac{1}{2}$

4 2

Ans.
(4)

Area of $△ O A B = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Now, Area of $△$ AMN = $\frac{4}{9}$ Area of $△$ OAB

$= \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$ ... (i)

Since, $∠$ OAB = 45°, then let $∠$ MAN = $θ$ and $∠$ MAO = 45° – $θ$ , then AM = sec (45° – $θ$ );

$A N = A M \times \frac{A N}{A M} = \sec (45 ° - θ) \cos θ$

and $M N = A M \times \frac{M N}{A M} = \sec (45 ° - θ) \sin θ$

Now, $Area of △ A M N = \frac{1}{2} \times A N \times M N$

$= \frac{1}{2} \times \sec^{2} (45 ° - θ) \sin θ \cos θ = \frac{2}{9} [From (i)]$

$\Rightarrow \frac{\sin 2 θ}{1 + \sin 2 θ} = \frac{4}{9} \Rightarrow \sin 2 θ = \frac{4}{5}$

$\Rightarrow 2 \tan^{2} θ - 5 \tan θ + 2 = 0$

$\Rightarrow \tan θ = \frac{1}{2}, 2 (Rejected as θ < 45 °)$

$∵ ∠ O B A = 45 °$

$∴ \tan 45 ° = \frac{M N}{B N} \Rightarrow M N = B N$

Now, $In △ A M N, \cot θ = \frac{A N}{M N} = \frac{A N}{B N} = \frac{λ}{1} \Rightarrow λ = 2$ $[∵ \tan θ = \frac{1}{2}]$

Hence, required sum is 2.

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