The line x + y = a meets the axes of x and y at A and B respectively. A triangle AMN is inscribed in the ΔOAB, O being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of the ΔAMN is 38 of the area of the ΔOAB, then ANBN is equal to
Let ANBN=λ.Then,the coordinates of N are(a1+λ,λa1+λ)
Where (a, 0) and (0, a) are the coordinates of A and B respectively. Now, equation of MN perpendicular to AB is
y−λa1+λ=x−a1+λ⇒x−y=1−λ1+λaSo,the coordinates of M are(0,λ−1λ+1a)=12∣∣∣[a(−aλ+1)+1−λ(1−λ)2a2]∣∣∣Also,area of the ΔOAB=a22So,that according to the given condition.λa2(1+λ)2=38.12a2⇒3λ2−10λ+3=0⇒λ=3⇒λ=13or 13Forλ=13,M lies outside the segment OB and hence the required value of λ is 3.Hence,(d)is the correct answer.