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 $\frac{AN}{BN}=\lambda$

Then, coordinate of N is $\left(\frac{a}{1+\lambda },\frac{a\lambda }{1+\lambda }\right)$

$\because SlopeofAB=-1$

$\therefore SlopeofMN=1$

$\therefore$ equation of MN is$y-\frac{a\lambda }{1+\lambda }=x-\frac{a}{1+\lambda }$

                      $\Rightarrow x-y=a\left(\frac{1-\lambda }{1+\lambda }\right)$

So, the coordinates of M are $\left(0,a\left(\frac{\lambda -1}{\lambda +1}\right)\right)$

Therefore, area of $\Delta AMN=\frac{3}{8}areaof\Delta OAB$

               $\Rightarrow \frac{1}{2}.AN.MN=\frac{3}{8}.\frac{1}{2}a.a$

             $\Rightarrow \frac{1}{2}.\left|\frac{a\lambda \sqrt{2}}{1+\lambda }.\frac{a\sqrt{2}}{1+\lambda }\right|=\frac{3}{8}.\frac{1}{2}a.a$

$\Rightarrow \frac{a^2\lambda }{{(1+\lambda )}^2}=\frac{3}{8}.\frac{1}{2}{a}^2$

$\therefore \lambda =3or\lambda =1/3$

For$\lambda =1/3$ , then M lies outside the segment OB and hence the required value of $\lambda =3$ .