Question

A variable line $L$ is drawn through $O(0,0)$ to meet the lines $L_1:y-x-10=0$ and $L_2:y-x-20=0$ 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 $O$. The locus of $P$ is

• A. $3x+3y=40$
• B. $3x+3y+40=0$
• C. $3x-3y=40$
• D. $3y-3x=40$

Answer 1

Answer: (D)

$3y-3x=40$

$x=r\cos \theta ,y=r\sin \theta$

$\therefore$ putting in $L_1$, we get

$\frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10}$

Similarly, putting in $L_2$, we get

$\frac{1}{OB}=\frac{\sin \theta -\cos \theta }{20}$

$\Rightarrow h=r\cos \theta ,k=r\sin \theta$

$\therefore \frac{2}{r}=\left(\frac{\sin \theta -\cos \theta }{10}\right)+\left(\frac{\sin \theta -\cos \theta }{20}\right)$

$\Rightarrow 40=3\left(r\sin \theta \right)-3\left(r\cos \theta \right)$

$\Rightarrow 3y-3x=40$