question_answer

The line \[x+y=a\] meets the axes of x and y at A and B respectively. A \[\Delta AMN\] 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) \[\frac{1}{3}\]

B) \[\frac{1}{3},3\]

C) \[\frac{2}{3},3\]

D) 3

Correct Answer: D

Solution :

[d] Let \[\frac{AN}{BN}=\lambda .\] Then, the coordinates of N are \[\left( \frac{a}{1+\lambda },\frac{\lambda a}{1+\lambda } \right).\]

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

\[y-\frac{\lambda a}{1+\lambda }=x-\frac{a}{1+\lambda }\]

\[\Rightarrow x-y=\frac{1-\lambda }{1+\lambda }a\]

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

Therefore, area of the \[\Delta AMN\]is

\[=\frac{1}{2}\left| \left[ a\left( \frac{-a}{\lambda +1} \right)+\frac{1-\lambda }{{{(1+\lambda )}^{2}}}{{a}^{2}} \right] \right|\]

\[=\frac{\lambda {{a}^{2}}}{{{(1+\lambda )}^{2}}}\]

Also, area of \[\Delta OAB=\frac{{{a}^{2}}}{2}\]

So, that according to the given condition

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

\[\Rightarrow 3{{\lambda }^{2}}-10\lambda +3=0\]

\[\Rightarrow \lambda =3\,\,or\,\,\lambda =\frac{1}{3}\]

For \[\lambda =\frac{1}{3},M\] lies outside the segment OB and

Hence the required value of \[\lambda \]is 3.