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The line $\mathrm{x+y=p}$ meets the axis of $\mathrm{x\, \& \, y}$ at $\mathrm{A\, \& \, B}$ respectively . A triangle APQ is inscribed in the triangle $\mathrm{OAB}, \mathrm{O}$ being the origin, with right angle at $\mathrm{Q}$ . $\mathrm{P}$ and $\mathrm{Q}$ lie respectively on $\mathrm{OB}$ and $\mathrm{AB}$ .If the area of the triangle $\mathrm{APQ}$ is $3 / 8^{\text {th }}$ of the area of the triangle $\mathrm{OAB}$ , then $\frac{\mathrm{AQ}}{\mathrm{BQ}}$ is equal to :

Option: 1
$2$

Option: 2
$2/3$

Option: 3
$1/3$

Option: 4
$3$

Answers (1)

$\frac{\triangle \mathrm{AQP}}{\triangle \mathrm{AOB}}=\frac{3}{8}$ or $\frac{\mathrm{p}^{2} \lambda}{\frac{(\lambda+1)^{2}}{2} \mathrm{p}^{2}}=\frac{3}{8}$
$\Rightarrow \quad \lambda=3, \frac{1}{3}$

$\frac{\mathrm{AQ}}{\mathrm{BQ}}=3$ or $\frac{1}{3}$
$\frac{1}{3}$ is rejected because this gives negative coordinater of $\mathrm{P}$ and it is gives that $\mathrm{P}$ lies on $\mathrm{OB}$ .

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Kshitij

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