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 The line \mathrm{x + y = a}, meets the axis of \mathrm{x} and y at \mathrm{A} and \mathrm{B} respectively. \mathrm{A} triangle \mathrm{AMN} is inscribed in the triangle \mathrm{OAB}, \mathrm{O} being the origin, with right angle at \mathrm{N.M} and \mathrm{N} lie respectively on \mathrm{OB} and \mathrm{AB}. If the area of the triangle \mathrm{AMN} is \mathrm{3/8} of the area of the triangle \mathrm{OAB}, then \mathrm{AN/BN=} 

 

Option: 1

\mathrm{1}


Option: 2

\mathrm{\frac{1}{2}}


Option: 3

\mathrm{3}


Option: 4

\mathrm{4}


Answers (1)

best_answer

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

 

 

 

 

Posted by

Suraj Bhandari

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