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The line \mathrm{x}+\mathrm{y}=\mathrm{p} meets the axis of \mathrm{x} \, \&\, \mathrm{y} at A & B respectively. A triangle \mathrm{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{A P Q} is \mathrm{3 / 8^{\text {th }}} of the area of the triangle \mathrm{O A B}, then \frac{A Q}{B Q} is equal to :

Option: 1

2


Option: 2

2 / 3


Option: 3

1 / 3


Option: 4

3


Answers (1)

best_answer

\frac{\triangle \mathrm{AQP}}{\Delta \mathrm{AOB}}=\frac{3}{8} \text { or } \frac{\frac{\mathrm{p}^2 \lambda}{(\lambda+1)^2}}{\frac{1}{2} \mathrm{p}^2}=\frac{3}{8}

\Rightarrow \quad \lambda=3, \frac{1}{3}
\frac{\mathrm{AQ}}{\mathrm{BQ}}=3 \text { or } \frac{1}{3}

\frac{1}{3}  is rejected because this gives negative coordinater of P and it is gives that P lies on OB.

Posted by

Rakesh

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