question_answer

A variable line L is drawn through \[O\left( 0,0 \right)\]to meet the lines \[{{L}_{1}}:y-x-10=0\]and \[{{L}_{2}}:\text{ }y-\text{ }x-20=0\] at the points A and B respectively. A point P is taken on L such that -\[\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}\]and P, A, B lies on same side of origin 0. The locus of P is

A)  \[3x+3y=40\]   

B)  \[3x+3y+40=0\]

C)  \[3x-3y=40\]    

D)  \[3y-3x=40\]

Correct Answer: D

Solution :

\[x=r\,\cos \,\theta ,\,\,y=r\sin \theta \]             \[\therefore \]putting in \[{{L}_{1}}\], we get                         \[\frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10}\] Similarly, putting in \[{{L}_{2}}\], we get \[\frac{1}{OB}=\frac{\sin \,\theta -\cos \theta }{20}\] \[\Rightarrow \,h=r\cos \theta ,k=r\sin \theta \] \[\therefore \,\frac{2}{r}=\left( \frac{\sin \theta -\cos \theta }{10} \right)+\left( \frac{\sin \theta -\cos \theta }{20} \right)\] \[\Rightarrow 40=3(r\,\sin \,\theta )-3(r\,\cos \,\theta )\,\Rightarrow \,3y-3x=40\]