Question

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

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

Answer 1

Answer: (A)

Let $\frac{A N}{B N}=\lambda .$
Then $N \equiv\left(\frac{a}{1+\lambda}, \frac{a \lambda}{1+\lambda}\right)$
Slope of $M N$ is 1 . So, equation of $M N$ is
$y-\frac{a \lambda}{1+\lambda}=x-\frac{a}{1+\lambda} \dots$(i)
Hence, $M$ is $\left(0, \frac{a(\lambda-1)}{\lambda+1}\right)$

$=\frac{1}{2} A N \times M N$
$=\frac{1}{2}\left|\frac{\sqrt{2} a \lambda}{1+\lambda} \times \frac{\sqrt{2} a}{1+\lambda}\right|$
$=\frac{a^{2} \lambda}{(\lambda+1)^{2}}$
Area of $\Delta O A B=\frac{1}{2} a^{2}$.
According to given condition:
$\frac{a^{2} \lambda}{(\lambda+1)^{2}}=\frac{3}{8} \times \frac{1}{2}$
$a^{2} \Rightarrow 3 \lambda^{2}-10 \lambda+3=0 $
$\Rightarrow \lambda=3, \frac{1}{3}$
For $\lambda=\frac{1}{3}, M$
lies outside the segment $O B$,
hence $\lambda=3 .$