Question

The line $x+y=a$ meets the $x$ -axis of $A$ and $y$ -axis at $B$. $A$ 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 area of $\Delta AMN$ is $\frac{3}{8}$ of the area of $\Delta OAB$, then $\frac{AN}{BN}$ is equal to

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

Answer 1

Answer: (A)

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

$=\frac{1}{2}AN\times MN$
$=\frac{1}{2}\left|\frac{\sqrt{2}a\lambda }{1+\lambda }\times \frac{\sqrt{2}a}{1+\lambda }\right|$
$=\frac{a^2\lambda }{(\lambda +1{)}^2}$
Area of $\Delta OAB=\frac{1}{2}{a}^2$.
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 $OB$,
hence $\lambda =3.$