Question

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 $\frac{3}{8}$ of the area of the triangle $OAB$, then $\frac{AN}{BN}$ is equal to

• A. 3
• B. $\frac{1}{3}$
• C. 2
• D. $\frac{1}{2}$

Answer 1

The correct option is A

3

Let $\frac{AN}{BN}=\lambda$. Then the coordinates of N are $(\frac{a}{1+\lambda },\frac{\lambda a}{1+\lambda })$
where $(a,0)$ and $(0,a)$ are the coordinates of $A$ and $B$ respectively.
Now equation of $MN$ perpendicular to $AB$ is
$y-\frac{\lambda a}{1+\lambda }=x-\frac{a}{1+\lambda }$
or $x-y=\frac{1-\lambda }{1+\lambda }a$
So, the coordiantes of M are
$(0,\frac{\lambda -1}{\lambda +1}a)$
Therefore, area of the triangle $AMN$ is
$=\frac{1}{2}\left|[a\left(\frac{-a}{\lambda +1}\right)+\frac{1-\lambda }{(1+\lambda {)}^2}{a}^2]\right|=\frac{\lambda a^2}{(1+\lambda {)}^2}$
Also area of the triangle $OAB$ = $\frac{a^2}{2}$
So that according to the given condition.
$\frac{\lambda a^2}{(1+\lambda {)}^2}=\frac{3}{8}.\frac{1}{2}{a}^2\Rightarrow 3{\lambda }^2-10\lambda +3=0\Rightarrow \lambda =3or\lambda =\frac{1}{3}$
For $\lambda =\frac{1}{3}$, M lies outside the segment OB and hence the required value of $\lambda$ is 3.