Question

A variable line $L$ is drawn through $O(0,0)$ to meet $L_1:x-y-8=0$ and $L_2:x-y-16=0$ at points $A$ and $B$ respectively. A point $P$ is taken on $L$ such that $\frac{1}{4OP}=\frac{1}{OA}+\frac{1}{OB}$. Then the locus of $P$ is

• A. $3x-3y=4$
• B. $3x+3y-4=0$
• C. $3x-3y=16$
• D. $x=y$

Answer 1

The correct option is A $3x-3y=4$

Line $L$ is passing through $(0,0)$
So, let equation of the line $L$ be $y=mx$.
Any point on the line $L$ is $(x,mx)$.

Finding intersection point $A$ of $L$ and $L_1:$
$x-mx=8\Rightarrow x=\frac{8}{1-m},y=\frac{8m}{1-m}$
Coordinates of $A$ is $(\frac{8}{1-m},\frac{8m}{1-m})$

Similarly, coordinates of $B$ is $(\frac{16}{1-m},\frac{16m}{1-m})$

Now, $\frac{1}{4OP}=\frac{1}{OA}+\frac{1}{OB}$
$\Rightarrow \frac{1}{\sqrt{{\left(\frac{8}{1-m}\right)}^2+{\left(\frac{8m}{1-m}\right)}^2}}+\frac{1}{\sqrt{{\left(\frac{16}{1-m}\right)}^2+{\left(\frac{16m}{1-m}\right)}^2}}=\frac{1}{4\sqrt{x^2+(mx{)}^2}}\Rightarrow \frac{1-m}{8}+\frac{1-m}{16}=\frac{1}{4x}$
$\Rightarrow 3x-3mx=4$
$\Rightarrow 3x-3y=4$