Question

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

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

Answer 1

Answer: (D)

Let $\frac{A N}{B \lambda}=\lambda$

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 $A B=\frac{0-a}{a-0}=-1$
$\therefore $ Slope of $M N=1(\because M N \perp A B)$
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 $\Delta A M N=\frac{3}{8} .$ area of $\Delta O A B$
$\Rightarrow \frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ a & 0 & \frac{a}{1+\lambda} \\ 0 & \left(\frac{\lambda-1}{\lambda+1}\right) & \frac{\lambda a}{1+\lambda}\end{vmatrix}=\frac{3}{8} \cdot \frac{1}{2} a^{2}$
$C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$
$\Rightarrow \frac{1}{2}\begin{vmatrix}1 & 0 & 0 \\ a & -a & \frac{-a \lambda}{1+\lambda} \\ 0 & \left(\frac{\lambda-1}{\lambda+1}\right) & \frac{\lambda a}{1+\lambda}\end{vmatrix}=\frac{3}{16} a^{2}$
$\Rightarrow \frac{1}{2}\left|-a\left(\frac{\lambda a}{1+\lambda}\right)+\frac{a \lambda}{1+\lambda}\left(\frac{\lambda-1}{\lambda+1}\right) a\right|=\frac{3}{16} a^{2}$
$\Rightarrow \frac{1}{2} \lambda a^{2}\left|\frac{\lambda-1}{(\lambda+1)^{2}}-\frac{1}{\lambda+1}\right|=\frac{3}{16} a^{2}$
$\Rightarrow \frac{1}{2} \lambda a^{2}\left|\frac{-2}{(\lambda+1)^{2}}\right|=\frac{3}{16} a^{2}$
$\Rightarrow \frac{\lambda a^{2}}{(\lambda+1)^{2}}=\frac{3}{16} a^{2}$
$\Rightarrow 16 \lambda=3 \lambda^{2}+3+6 \lambda$
$\Rightarrow 3 \lambda^{2}-10 \lambda+3=0$
$\Rightarrow (\lambda-3)(3 \lambda-1)=0$
$\Rightarrow \lambda=3$ 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, $\frac{A N}{B N}=3$