Question

A line through $P(3,4)$ cuts the lines $x=6$ and $y=8$ at $L$ and $M$ respectively. $Q$ is a variable point on the line such that $\frac{1}{PQ}=\frac{1}{PL}+\frac{1}{PM}$ then the locus of $Q$ is

• A. $4x+3y-36=0$
• B. $x^2+y^2=36$
• C. $3x-4y-36=0$
• D. $4x^2-9y^2=36$

Answer 1

The correct option is A $4x+3y-36=0$

Let inclination of line passing through point $(3,4)$ be $\theta$
So, parametric coordinates of any point on this line will be $(3+r\cos \theta ,4+r\sin \theta )$
Substituting this in $y=8$, we get $PL=\frac{4}{\sin \theta }$
Substituting this in $x=6$ we get $PM=\frac{3}{\cos \theta }$
Let locus of $Q$ be $(h,k)=(3+r\cos \theta ,4+r\sin \theta ),r=PQ$
$\frac{1}{PQ}=\frac{1}{PL}+\frac{1}{PM}$
$\Rightarrow \frac{1}{PQ}=\frac{\sin \theta }{4}+\frac{\cos \theta }{3}\Rightarrow 12=3r\sin \theta +4r\cos \theta$
$12=3(k-4)+4(h-3)$
$\Rightarrow 3k-12+4h-12-12=0$
$\therefore$ required locus equaton is $4x+3y-36=0$