Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$Let\frac{AN}{BN}=\lambda .Then,thecoordinatesofNare(\frac{a}{1+\lambda },\frac{\lambda a}{1+\lambda })$

Where (a, 0) and (0, a) are the coordinates of A and B respectively. Now, equation of MN perpendicular to AB is

$y-\frac{\lambda a}{1+\lambda }=x-\frac{a}{1+\lambda }\Rightarrow x-y=\frac{1-\lambda }{1+\lambda }aSo,thecoordinatesofMare(0,\frac{\lambda -1}{\lambda +1}a)=\frac{1}{2}\left|[a\left(\frac{-a}{\lambda +1}\right)+\frac{1-\lambda }{(1-\lambda {)}^2}{a}^2]\right|Also,areaofthe\Delta OAB=\frac{a^2}{2}So,thataccordingtothegivencondition.\frac{\lambda a^2}{(1+\lambda {)}^2}=\frac{3}{8}.\frac{1}{2}{a}^2\Rightarrow 3{\lambda }^2-10\lambda +3=0\Rightarrow \lambda =3\Rightarrow \lambda =\frac{1}{3}or\frac{1}{3}For\lambda =\frac{1}{3},MliesoutsidethesegmentOBandhencetherequiredvalueof\lambda is3.Hence,\left(d\right)isthecorrectanswer.$