Question

The line $x+y=a$ meets the axes of $X$ and $Y$ at $A$ and $B$ respectively. $A\Delta AMN$ is inscribed in the $\Delta O$ A B, O being the origin, with right angle at N. M and $N$ lie on OB and $AB$ respectively. 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. $3$

Answer 1

Answer: (D)

Let $\frac{AN}{B\lambda }=\lambda$

then by section formula, 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.
Since, slope of $AB=\frac{0-a}{a-0}=-1$
$\therefore$ Slope of $MN=1(\because MN\perp AB)$
Now, equation of MN
$y-\frac{\lambda a}{1+\lambda }=x-\frac{a}{1+\lambda }$
$\Rightarrow x-y=\frac{1-\lambda }{1+\lambda }a$
By putting $x=0$ in this equation, we get the
coordinates of $M$ which are $\left[0,\left(\frac{\lambda -1}{\lambda +1}\right)a\right]$.
Now, it is given that
area of $\Delta AMN=\frac{3}{8}.$ area of $\Delta OAB$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ a& 0& \frac{a}{1+\lambda }\\ 0& \left(\frac{\lambda -1}{\lambda +1}\right)& \frac{\lambda a}{1+\lambda }\end{array}\right|=\frac{3}{8}\cdot \frac{1}{2}{a}^2$
$C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}1& 0& 0\\ a& -a& \frac{-a\lambda }{1+\lambda }\\ 0& \left(\frac{\lambda -1}{\lambda +1}\right)& \frac{\lambda a}{1+\lambda }\end{array}\right|=\frac{3}{16}{a}^2$
$\Rightarrow \frac{1}{2}\left|-a\left(\frac{\lambda a}{1+\lambda }\right)+\frac{a\lambda }{1+\lambda }\left(\frac{\lambda -1}{\lambda +1}\right)a\right|=\frac{3}{16}{a}^2$
$\Rightarrow \frac{1}{2}\lambda a^2\left|\frac{\lambda -1}{(\lambda +1{)}^2}-\frac{1}{\lambda +1}\right|=\frac{3}{16}{a}^2$
$\Rightarrow \frac{1}{2}\lambda a^2\left|\frac{-2}{(\lambda +1{)}^2}\right|=\frac{3}{16}{a}^2$
$\Rightarrow \frac{\lambda a^2}{(\lambda +1{)}^2}=\frac{3}{16}{a}^2$
$\Rightarrow 16\lambda =3{\lambda }^2+3+6\lambda$
$\Rightarrow 3{\lambda }^2-10\lambda +3=0$
$\Rightarrow (\lambda -3)(3\lambda -1)=0$
$\Rightarrow \lambda =3$ 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 .
ie, $\frac{AN}{BN}=3$