The line x+y=a meets the axis of x and y at A and B respectively a triangle AMN is inscribed in the triangle OAB, O being the origin, with right angle at N,M and N lie respectively on OB and AB. If the area of the triangle AMN is 3/8 of the area of the triangle OAB, then AN/BN is equal to

Let ANBN=λ Then, coordinate of N is (a1+λ,aλ1+λ) ∵      Slope  of  AB=−1                  ∴      Slope  of  MN=1 ∴ equation of MN isy−aλ1+λ=x−a1+λ                       ⇒                  x−y=a(1−λ1+λ) So, the coordinates of M are (0,a(λ−1λ+1)) Therefore, area of ΔAMN=38area  of  ΔOAB                ⇒       12.AN.MN=38.12a.a              ⇒       12.|aλ21+λ.a21+λ|=38.12a.a ⇒       a2λ(1+λ)2=38.12a2 ∴           λ=3  or  λ=1/3 Forλ=1/3 , then M lies outside the segment OB and hence the required value of λ=3 .