Question

From a point $P$ perpendiculars $PM$ and $PN$ are drawn upon two fixed lines which are inclined at an angle $\omega$, and which are taken as the axes of coordinates and meet in $O$; find the locus of $P$ if $MN$ be parallel to the given line $y=mx$.

Answer 1

Let the point $P$ be $(h,k)$

Draw $P{L}^{\prime }$ parallel to $MO$ and $PL$ parallel to $NO$

In $△PLM\cos \omega =\frac{LM}{PL}$

$OM=OL+LMOM=h+k\cos \omega$

So, the coordinates of $M$ are $(h+k\cos \omega ,0)$

In $△P{L}^{\prime }N\cos \omega =\frac{L^{\prime }N}{P{L}^{\prime }}$

$\Rightarrow L^{\prime }N=P{L}^{\prime }\cos \omega =h\cos \omega ON=O{L}^{\prime }+L^{\prime }NON=k+h\cos \omega$

So, the coordinates of $N$ are $(0,k+h\cos \omega )$

Slope of $MN$ $=\frac{k+h\cos \omega -0}{0-h+k\cos \omega }=-\frac{k+h\cos \omega }{h+k\cos \omega }$

$MN$ is parallel to $y=mx$

$\Rightarrow -\frac{k+h\cos \omega }{h+k\cos \omega }=mk+h\cos \omega =-mh-mk\cos \omega k+mk\cos \omega +h\cos \omega +mh=0k(1+m\cos \omega )+h(m+\cos \omega )=0$

Replacing $h$ by $x$ and $y$ by $k$

$y(1+m\cos \omega )+x(m+\cos \omega )=0$

is the required locus of $P$.